Evaluate the following integrals:
Formula =
Therefore ,
Put 2x + 9 = t ⇒ 2 dx = dt
Evaluate the following integrals:
Formula =
Therefore ,
Put 7 – 3x = t ⇒ -3 dx = dt
Evaluate the following integrals:
Formula =
Therefore ,
Put 3x - 5 = t ⇒ 3 dx = dt
Evaluate the following integrals:
Formula =
Therefore ,
Put 4x + 3 = t ⇒ 4 dx = dt
Evaluate the following integrals:
Formula =
Therefore ,
Put 3 – 4x = t ⇒ -4 dx = dt
Evaluate the following integrals:
Formula =
Therefore ,
Put 2x – 3 = t ⇒ 2 dx = dt
Formula =
Therefore ,
Put 2x – 1 = t ⇒ 2 dx = dt
Evaluate the following integrals:
Formula =
Therefore ,
Put 1 – 3x = t ⇒ -3 dx = dt
Evaluate the following integrals:
Formula =
Therefore ,
Put 2 – 3x = t ⇒ -3 dx = dt
Evaluate the following integrals:
Formula =
Therefore ,
Put 3x = t ⇒ 3 dx = dt
Evaluate the following integrals:
Formula =
Therefore ,
Put 5 + 6x = t ⇒ 6 dx = dt
Evaluate the following integrals:
Formula
Therefore ,
Put sin x =t ⇒ cos x dx = dt
Evaluate the following integrals:
Formula
Therefore ,
Put 2x + 5 =t ⇒ 2 dx = dt
Evaluate the following integrals:
Formula
Therefore ,
Put sin x =t ⇒ cos x dx = dt
Evaluate the following integrals:
Formula
Therefore ,
Put sin x =t ⇒ cos x dx = dt
Evaluate the following integrals:
Formula
Therefore ,
Put cos x =t ⇒ -sin x dx = dt
Evaluate the following integrals:
Formula
Therefore ,
Put =t ⇒ dx = dt
Evaluate the following integrals:
..
Formula
Therefore ,
Put ⇒
Evaluate the following integrals:
Formula
Therefore ,
Put ⇒
Evaluate the following integrals:
Formula
Therefore ,
Put ⇒
Evaluate the following integrals:
Formula
Therefore ,
Put ⇒
Evaluate the following integrals:
Formula
Therefore ,
Put ⇒
Evaluate the following integrals:
Formula
Therefore ,
Put ⇒
Evaluate the following integrals:
Formula
Therefore ,
Put ⇒
Evaluate the following integrals:
Formula =
Therefore ,
Put tan x = t ⇒ dx = dt
Evaluate the following integrals:
Formula =
Therefore ,
Put = t ⇒ dx = dt
Evaluate the following integrals:
Formula =
Therefore ,
Put ax+b = t ⇒ adx = dt
Put sin t = z � cos t dt = dz
Evaluate the following integrals:
Formula =
Therefore ,
Evaluate the following integrals:
Formula =
Therefore ,
Put = t ⇒
Evaluate the following integrals:
Formula =
Therefore ,
Put = t ⇒
Evaluate the following integrals:
Formula =
Therefore ,
Put = t ⇒
Evaluate the following integrals:
Formula =
Therefore ,
Put = t ⇒
Evaluate the following integrals:
Formula =
Therefore ,
Put log (sin x) = t ⇒ �
Evaluate the following integrals:
Formula =
Therefore ,
Put log (sin x) = t ⇒ �
Evaluate the following integrals:
Formula =
Therefore ,
Put x2 + 1= t ⇒
Evaluate the following integrals:
Formula =
Therefore ,
Put log (sec x + tan x)= t
Sec x dx = dt
Evaluate the following integrals:
Formula =
Therefore ,
= t
Evaluate the following integrals:
Formula =
Therefore ,
Put = t ⇒ �
Evaluate the following integrals:
Formula =
Therefore ,
Put = t ⇒ �
Evaluate the following integrals:
Formula =
Therefore ,
Put = t ⇒ �
Evaluate the following integrals:
Formula =
Therefore ,
Put 2 + log x = t ⇒
Evaluate the following integrals:
Formula =
Therefore ,
Put 1 + tan x = t ⇒
Evaluate the following integrals:
Formula =
Therefore ,
Put 1 + cos x = t ⇒
Evaluate the following integrals:
Formula =
Therefore ,
Put cos x - sin x = t ⇒ (- cos x - sin x) dx = dt
Evaluate the following integrals:
i.
ii.
(i)
Formula =
Therefore ,
Put x + log (sec x) = t ⇒
(ii)
Formula =
Therefore ,
Put = t ⇒
Evaluate the following integrals:
Formula =
Therefore ,
Put �
Evaluate the following integrals:
Formula =
Therefore ,
Put
Evaluate the following integrals:
Formula =
Therefore ,
Put 3cos x + 2sin x = t ⇒ (2cos x - 3sin x) dx = dt
Evaluate the following integrals:
Formula =
Therefore ,
Put 2x2 +3= t ⇒ (4x) dx = dt
Evaluate the following integrals:
Formula =
Therefore ,
Put x2+2x+3= t ⇒ (2x+2) dx = dt � 2(x+1)dx=dt
To find: Value of
Formula used:
We have, … (i)
Let 2x2 - 5x + 1 = t
⇒ (4x - 5)dx = dt
Putting this value in equation (i)
Ans)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let 3x3 - 2x2 + 5x + 1 = t
⇒ (9x2 - 4x + 5)dx = dt
Putting this value in equation (i)
Ans)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let = t
⇒ ()dx = dt
Putting this value in equation (i)
Ans)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let t
⇒ ()dx = dt
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let t
⇒ ()dx = dt
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let = t
⇒ () dx = dt
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let t
⇒ () dx = dt
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let t
⇒ t - 1
⇒ dx =
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let t
⇒ t - 1
⇒ dx =
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Multiplying numerator and denominator with x
Let t
⇒ xdx =
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let t
x = t + 1
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let t
x = t - 1
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let = t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let = t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let = t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let = t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let = t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let = t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let = t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let = t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let = t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used: (i)
(ii)
We have, … (i)
Let = t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let = t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let = t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let = t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let = t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let = t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let = t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let = t2
= t2
t2 + 1
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let = t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let = t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let = t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let = t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let = t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let = t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let = t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used: (i)
(ii)
We have, … (i)
Let
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let = t , x = sint ,
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let = t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Dividing Numerator and Denominator by x2,
Let = t
Putting this value in equation (i)
Evaluate the following integrals:
To find: Value of
Formula used:
We have, … (i)
Let (sinx – cosx) = t
⇒ t2 = sin2x - 2sinx. cosx + cos2x
⇒ t2 = 1 - 2sinx.cosx
⇒ 2sinx.cosx = 1 - t2
⇒ sin2x = 1 - t2
Putting this value in equation (i)
Let
⇒ … (ii)
Now if
Then cosθ =
⇒ cosθ =
⇒ cosθ =
⇒ cosθ =
Now tanθ =
Now tanθ =
Comparing the value θ from eqn. (ii)
Dividing Numerator and denominator from cosx
Ans.)
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Given =
Let, 2x + 3 = z
⇒ 2dx = dz
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A. -5(3 – 5x)6 + C
B.
C.
D. none of these
Given =
Let, 3 – 5x = z
⇒ -5dx = dz
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Given =
Let, 2 – 3x = z
⇒ -3dx = dz
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Given =
Let, ax + b = z2
⇒ adx = 2zdz
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B.
C. 4 tan (7 – 4x) + C
D. - 4 tan (7 – 4x) + C
Given =
Let, 7 – 4x = z
⇒ -4dx = dz
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B.
C. 3 sin 3x + C
D. -3 sin 3x + C
Given =
So, where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Given =
Let, 5 – 3x = z
⇒ -3dx = dz
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Given =
Let, 3x + 4 = z
⇒ 3dx = dz
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D.
Given =
Let,
⇒ dx = 2dz
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D.
Given =
So,
Let 1 + cosx = u2
So, -sinxdx = 2udu
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Given =
So,
Let 1 - sinx = u2
So, -cosxdx = 2udu
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Given =
So,
Let cosx = u
So, -sinxdx = du
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D.
Given =
Let, logx = u
So,
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A. log (tan x) + C
B. - log (tan x) + C
C. tan (tan x) + C
D. - tan (log x) + C
Given =
Let, logx = z
⇒
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B.
C. (log x)2 + C
D. log |log x| + C
Given =
Let, logx = z
⇒
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Given =
Let, x3 = z
⇒ 3x2dx = dz
⇒
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Given =
Let, x = z2
⇒ dx = 2zdz
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Given =
Let, tan-1x = z
⇒
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D.
Given =
Let, x = z2
⇒ dx = 2zdz
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D.
Given =
Let, sinx = z2
⇒ cosxdx = 2zdz
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Given =
Let, tan-1x = z2
⇒
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A. log |cot x| + C
B. log |cot x cosec x| + C
C. log |log sin x| + C
D. none of these
Given =
Let, sinx = z
⇒ cosxdx = dz
So,
Let, logz = u
⇒
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A. tan (1 + log x) + C
B. cot (1 + log x) + C
C. sec (1 + log x) + C
D. none of these
Given =
Let, 1 + logx = z
⇒
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B. log |tan-1 x3| + C
C.
D. none of these
Given =
Let, tan-1x3 = z
⇒
⇒
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A. 5 tan5 x + C
B.
C. 5 log |cos x| + C
D. none of these
Given =
So,
Let, secx = z
⇒ secxtanxdx = dz
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D.
Given =
So,
Let, cosec(2x + 1) = z
⇒ -2cosec(2x + 1)cot(2x + 1)dx = dz
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A. log |sec (sin-1 x)| + C
B. log |cos (sin-1 x)| + C
C. tan (sin1 x) + C
D. none of these
Given =
Let, sin-1x = z
⇒
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A. x tan (log x) + C
B. log |tan x| + C
C. log |cos (log x)| + C
D. - log |cos (log x)| + C
Given =
Let, logx = z
⇒
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A. cot (ex) + C
B. log |sin ex| + C
C. log |cosec ex| + C
D. none of these
Given =
Let, ex = z
⇒ exdx = dz
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Given =
Let, 1 + ex = z2
⇒ exdx = 2zdz
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D.
Given =
Let, 1 – x2 = z2
⇒ -2xdx = 2zdz
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A. tan (xex) + C
B. cot (xex) + C
C. exx tan x + C
D. none of these
Given =
Let, xex = z
⇒ ex(1 + x)dx = dz
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A. cot-1 (ex) + C
B. tan-1 (ex) + C
C. log |ex + 1| + C
D. none of these
Given =
Let, ex + 1 = z
⇒ exdx = dz
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A. sin-1 (2x) + C
B. (log e2) sin-1 (2x) + C
C. (log e2) cos-1 (2x) + C
D. log2 e) sin-1 (2x) + C
Given =
Let, 2x = z
⇒ 2x(log2)dx = dz
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A. log |ex – 1| + C
B. log |1 – e-x| + C
C. log |ex – 1| + C
D. none of these
Given =
Let, ex – 1 = z
⇒ exdx = dz
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Given =
Let,
⇒
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A. tan x + sec x + C
B. tan x – sec x + C
C.
D. none of these
Given
Let,
⇒
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A. x + tan x – sec x + C
B. x – tan x – sec x + C
C. x – tan x + sec x + C
D. none of these
Given
Let,
⇒
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A. - x + sec x – tan x + C
B. x + cos x – sin x + C
C. - log |1 – sin x| + C
D. none of these
Given
Let,
⇒
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Given
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B. log |x – sin x| + C
C.
D.
Given
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D.
Given
Let,
⇒
So,
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Given
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Given
[From Question no. 40] where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A. tan x – cot x + C
B. tan x + cot x + C
C. - tan x + cot x + C
D. none of these
Given
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A. tan x + x + C
B. tan x – x + C
C. - tan x + x + C
D. none of these
Given
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Given
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A. tan x + cot x + C
B. tan x – cot x + C
C. - tan x + cot x + C
D. none of these
Given
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A. cot x + tan x + C
B. - cot x + tan x + C
C. cot x – tan x + C
D. - cot x – tan x + C
Given
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A. sin x + x cos α + C
B. 2sin x + x cos α + C
C. 2 sin x + 2x cos α + C
D. none of these
Given
where c is the integrating constant.
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Formula :-
Therefore ,
⇒
⇒
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Formula :-
Therefore ,
⇒
⇒
⇒
(Multiply by in numerator and denominator)
⇒
⇒
Mark (√) against the correct answer in each of the following:
A. 2 tan x + x – 2sec x + C
B. 2 tan x – x + 2 sec x + C
C. 2 tan x – x – 2sec x + C
D. none of these
Formula :-
Therefore ,
⇒
⇒
⇒
⇒
⇒
⇒
Put cos x = t
Therefore -> sin x dx = - dt
⇒
⇒
⇒
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Formula :-
Therefore ,
⇒
⇒
⇒
⇒
Mark (√) against the correct answer in each of the following:
A. x cos 2α - sin 2α . log |sin (x + α)| + C
B. x cos 2α + sin 2α . log|sin (x + α)| + C
C. x cos 2α + sin α . log |sin (x + α)| + C
D. none of these
Formula :-
Therefore ,
⇒
⇒
⇒
⇒
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Formula :-
Therefore ,
⇒ (Rationalizing the denominator)
⇒
⇒
⇒
Mark (√) against the correct answer in each of the following:
A. - log |cos x – sin x| + C
B. log |cos x – sin x| + C
C. log |cos x + sin x | + C
D. none of these
Formula :-
Therefore ,
⇒ (Rationalizing the denominator)
⇒
Put cos x - sin x = t
(- sin x - cos x) dx = dt
(sin x + cos x) dx = -dt
⇒
⇒
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Formula :-
Therefore ,
Put
⇒
⇒
⇒
⇒
Mark (√) against the correct answer in each of the following:
A. sin-1 x3 + C
B. cos-1 x3 + C
C. tan-1 x3 + C
D. cot-1 x3 + C
Formula :-
Therefore ,
Put
⇒
⇒
⇒
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Formula :-
Therefore ,
Put
⇒
⇒
⇒
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Formula :-
Therefore ,
⇒ (Rationalizing the denominator)
⇒
⇒
⇒
⇒
Mark (√) against the correct answer in each of the following:
A. log |cos x| + C
B. - log |cos x| + C
C. log |sin x| + C
D. - log |sin x| + C
Formula :-
Therefore ,
⇒
Put cos x = t -sin x dx = dt
⇒
⇒
⇒
Mark (√) against the correct answer in each of the following:
A. log |sec x - tan x| + C
B. - log |sec x + tan x| + C
C. log |sec x + tan x| + C
D. none of these
Formula :-
Therefore ,
⇒
⇒
Put , (dx = dt
⇒
⇒
Mark (√) against the correct answer in each of the following:
A. log |cosec x - cot x| + C
B. - log |cosec x - cot x| + C
C. log |cosec x + cot x| + C
D. none of these
Formula :-
Therefore ,
Put , dx = dt
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Formula :-
Therefore ,
Mark (√) against the correct answer in each of the following:
A. tan-1 (cos x) + C
B. - tan-1 (cos x) + C
C. cot-1 (cos x) + C
D. none of these
Formula :-
Therefore ,
Put sec x = t (sec x tan x) dx = dt
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Formula :-
Therefore ,
Put -2x dx = dt
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Formula :-
Therefore ,
Put
Mark (√) against the correct answer in each of the following:
A. tan-1 x4 + C
B. 4 tan-1 x4 + C
C.
D. none of these
Formula :-
Therefore ,
Put
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Formula :-
Therefore ,
Put ⇒
Mark (√) against the correct answer in each of the following:
A. (tan-1x2)2 + C
B. 2 tan-1 x2 + C
C.
D. none of these
Formula :-
Therefore ,
Put ⇒
Mark (√) against the correct answer in each of the following:
A. - 3 log |2 – 3x| + C
B.
C. - log |2 – 3x| + C
D. none of these
Formula :-
Therefore ,
Put
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Formula :-
Therefore ,
Put
⇒
Mark (√) against the correct answer in each of the following:
A.
B.
C. -3(5 – 3x) log 3 + C
D. none of these
Formula :-
Therefore ,
Put
⇒
Mark (√) against the correct answer in each of the following:
A. etan x + tan x + C
B. etan x . tan x + C
C. etan x + C
D. none of these
Formula :-
Therefore ,
Put
⇒
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Formula :-
Therefore ,
Put ⇒ ⇒
⇒
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Formula :-
Therefore ,
Put ⇒
⇒
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Formula :-
Therefore ,
Put ⇒
⇒
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Formula :-
Therefore ,
Put ⇒
⇒
Mark (√) against the correct answer in each of the following:
A. tan (xex) + C
B. - tan (xex) + C
C. cot (xex) + C
D. none of these
Formula :-
Therefore ,
Put ⇒ ⇒
⇒
Mark (√) against the correct answer in each of the following:
A. sec-1 x2 + C
B.
C. cosec-1 x2 + C
D. none of these
Formula :-
Therefore ,
Put ⇒
⇒
⇒
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Formula :-
Therefore ,
Put ⇒ x = t + 1 ⇒
⇒
⇒
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Formula :-
Therefore ,
Put ⇒
⇒
⇒
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Formula :-
Therefore ,
Put ⇒
⇒ ⇒ ⇒
⇒
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Formula :-
Therefore ,
Put ⇒
⇒
Put dt = 2z dz
⇒ ⇒
⇒
⇒
Mark (√) against the correct answer in each of the following:
A.
B.
C. log |sin x – cos x| + C
D. none of these
Formula :-
Therefore ,
We can write
⇒
Put
⇒ ⇒
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Formula :-
Therefore ,
⇒
We can write
⇒
Put
⇒ ⇒
Mark (√) against the correct answer in each of the following:
A. log |sin x – cos x| + C
B.
C.
D.
Formula :-
Therefore ,
⇒
We can write
⇒
Put
⇒ ⇒
Mark (√) against the correct answer in each of the following:
A. sin-1 (tan x) + C
B. cos-1 (sin x) + C
C. tan-1 (cos x) + C
D. tan-1 (sin x) + C
Formula :-
Therefore ,
Put ⇒
⇒
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Formula :-
Therefore ,
⇒ ⇒
Put ⇒
⇒
Mark (√) against the correct answer in each of the following:
A.
B.
C. log|cos6 x| + C
D. none of these
Formula :-
Therefore ,
⇒ ⇒
Put ⇒
⇒
Mark (√) against the correct answer in each of the following:
A.
B.
C. 5 log |cos x| + C
D. none of these
Formula :-
Therefore ,
Put ⇒
⇒
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Formula :-
Therefore ,
⇒
⇒
Put ⇒
⇒
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Formula :-
Therefore ,
⇒
⇒
Put ⇒
⇒
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Formula :-
Therefore ,
⇒
Put ⇒
⇒
Mark (√) against the correct answer in each of the following:
A. log {log (tan x )| + C
B.
C. log (sin x cos x) + C
D. none of these
Formula :-
Therefore ,
⇒
Put ⇒ ⇒
⇒
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Formula :-
Therefore ,
⇒
Put ⇒
⇒
⇒
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
Formula :-
Therefore ,
⇒ ⇒
Put ⇒
⇒ ⇒
Mark (√) against the correct answer in each of the following:
A. log |sin x – cos x| + C
B.
C. log |cos x + sin x| + C
D. none of these
Formula :-
Therefore ,
⇒
Put ⇒
⇒ ⇒
Mark (√) against the correct answer in each of the following:
A.
B. e
C.
D. none of these
Formula :-
Therefore ,
Put ⇒
⇒
Put dt = 2z dz
⇒ ⇒
⇒
⇒
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D.
Let
Now multiplying and dividing by cos2x, we get,
Let tan x = t
Differentiating both sides, we get,
sec2x dx = dt
Therefore,
Integrating, we get,
i)
⇒
Now, we know that 1-cos2x=2sin2x
So, applying this identity in the given integral, we get,
⇒
Ans:
ii)
⇒
Now, we know that 1+cos2x=2cos2x
So, applying this identity in the given integral, we get,
⇒
Ans:
Evaluate the following integrals:
(i)
(ii)
(i)
⇒
Now, we know that 1+cosx=2cos2 (x/2)
So, applying this identity in the given integral, we get,
⇒
Ans:
ii)
⇒
Now, we know that cosec2x-cot2 x=1
So, applying this identity in the given integral we get,
⇒
⇒
⇒
⇒-2cotx-x+c
⇒-2cotx-x+c
Ans: -2cotx-x+c
Evaluate the following integrals:
(i)
(ii)
i)
⇒
Now, we know that 1-cos2nx=2sin2nx
So, applying this identity in the given integral, we get,
⇒
Ans:
(ii)
We know that 1-cos2x=sin2x
⇒
⇒Put cosx=t
⇒-sinxdx=dt
⇒
⇒
⇒
⇒
Resubstituting the value of t=cosx we get,
⇒
Ans:
Evaluate the following integrals:
Substitute 3x+5=u
⇒3dx=du
⇒dx=du/3
⇒
Now We know that 1-cos2x=sin2x ,
⇒
⇒Substitute sinu=t
⇒cosu du=dt
⇒
⇒
⇒
⇒
Resubstituting the value of t=sinu and u=3x+5 we get,
⇒
Ans:
Evaluate the following integrals:
⇒
Substitute 2x-3=u
⇒ 2dx=du
⇒dx=du/2
⇒
⇒ We know that 1-cos2x=sin2x
⇒
⇒Put cosu=t
⇒-sinxdu=dt
⇒
⇒
⇒
⇒
⇒
Resubstituting the value of t=cosu and u=2x-3 we get
⇒
⇒
Now as we know cos(-x)=cosx
⇒
=
Ans:
Evaluate the following integrals:
(i)
(ii)
(i)
⇒
1-cos2x=2sin2x and 1+cos2x=2cos2x
⇒
⇒
Now sec2x-1=tan2x
⇒
⇒
⇒tanx-x+c
Ans: tanx-x+c
(ii)
⇒
1-cos2x=2sin2x and 1+cos2x=2cos2x
⇒
⇒
Now cosec2x-1=cot2x
⇒
⇒
⇒-cotx-x+c
Ans: -cotx-x+c
Evaluate the following integrals:
(i)
(ii)
i)
⇒
1-cosx=2sin2x/2 and 1+cosx=2cos2x/2
⇒
⇒
Now sec2(x/2)-1=tan2(x/2)
⇒
⇒
⇒2tan(x/2)-x+c
Ans: 2tan(x/2)-x+c
(ii)
⇒
1-cosx=2sin2x/2 and 1+cosx=2cos2x/2
⇒
⇒
Now cosec2 (x/2)-1=cot2 (x/2)
⇒
⇒
⇒-2cot(x/2)-x+c
Ans: ⇒-2cot(x/2)-x+c
Evaluate the following integrals:
⇒
Applying the formula: sinx×cosy=1/2(sin(x+y)-sin(y-x))
⇒
⇒
⇒
Ans:
Evaluate the following integrals:
⇒
Applying the formula: cosx×cosy=1/2(cos(x+y)+cos(x-y))
⇒
⇒
⇒
Ans:
Evaluate the following integrals:
⇒
Applying the formula: sinx×siny=1/2(cos(y-x)-cos(y+x))
⇒
⇒
⇒
Ans:
Evaluate the following integrals:
⇒
Applying the formula: sinx×cosy=1/2(sin(y+x)-sin(y-x))
⇒
⇒
⇒
Ans:
Evaluate the following integrals:
we know that 1+cos2x=2cos2x
So, applying this identity in the given integral we get,
⇒
⇒
⇒
Let sinx =t
⇒ cosx dx=dt
⇒
⇒
Resubstituting the value of t=sinx we get
⇒
Ans:
Evaluate the following integrals:
⇒
⇒ …()
⇒
⇒
⇒
⇒ …(1+cos4x=2cos2x)
⇒
⇒
⇒
⇒
Ans:
Evaluate the following integrals:
⇒
⇒
⇒
Ans:
Evaluate the following integrals:
Let sinx =t
⇒ cosx dx =dt
⇒
⇒
Resubstituting the value of t=sinx we get
Ans:
Evaluate the following integrals:
⇒
⇒
⇒Put tanx=t ⇒sec2dx=dt
⇒
⇒
Resubstituting the value of t=tanx we get
⇒
Ans:
Evaluate the following integrals:
⇒
⇒
⇒
Put sinx=t
⇒cosxdx=dt
⇒
⇒
⇒
Resubstituting the value of t=sinx we get,
⇒
Ans:
Evaluate the following integrals:
⇒
⇒
⇒
Put cosx=t
⇒-sinxdx=dt
⇒
⇒
⇒
Resubstituting the value of t=sinx we get,
⇒
Ans:
Evaluate the following integrals:
⇒
⇒
⇒
Put sinx=t
⇒cosxdx=dt
⇒
⇒
Resubstituting the value of t=sinx we get
Ans:
Evaluate the following integrals:
⇒
⇒
⇒
Put cosx=t
⇒-sinxdx=dt
⇒
⇒
Resubstituting the value of t=cosx we get
Ans:
Evaluate the following integrals:
cot2x=t ⇒-2cosec2 2xdx=dt
Resubstituting the value of t=cotx we get
Ans:
Evaluate the following integrals:
Ans: 2sinx-log|secx+tanx|+c
Evaluate the following integrals:
Now α is a constant
Ans:xcos α-sin αlog|cos(x+ α)|+c
Evaluate the following integrals:
Now put cosx=t
⇒-sinxdx=dt
Resubstituting the value of t= cosx we get,
Ans:
Evaluate the following integrals:
Put cosx =t
⇒ -sinxdx=dt
Now put t2-1=a
⇒2tdt=da
And t8=(a+1)4
Resubstituting the value of a=t2-1 and t=cosx ⇒a=cos2x-1=-sin2x we get
Ans:
Evaluate the following integrals:
⇒
⇒ …()
⇒
⇒
⇒
⇒ …(1+cos8x=2cos2 4x)
⇒
⇒
⇒
⇒
Ans:
Evaluate the following integrals:
Doing tangent half angle substitution we get,
Substitute u=tan(x/2)
⇒2du=sec2(x/2)dx
⇒dx=
⇒
⇒
⇒
⇒
Resubstituting the values we get,
Ans:
Evaluate the following integrals:
Let tan=t
∴
Resubstituting the value of t we get
Ans:
Evaluate the following integrals:
a > 0 and b > 0
Taking bcosx common from the denominator we get,
Let (a/b)+tanx=t
∴
Resubstituting the value of t = (a/b)+tanx we get
Ans:
Evaluate the following integrals:
Let tan=t
∴
resubstituting the value of t we get
Ans:
Evaluate the following integrals:
Ans: 4tanx-9cotx-25x+c
Evaluate the following integrals:
⇒
Applying the formula: sinx×siny=1/2(cos(y-x)-cos(y+x))
⇒
⇒
⇒
⇒
Ans:
Evaluate the following integrals:
Ans: -log(sinx+cosx)+c
Evaluate the following integrals:
Let tan=t
∴
Resubstituting the value of t we get
Ans:
Using BY PART METHOD.
Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here x is the first function and is the second function.
Using Integration by part
Evaluate the following integrals:
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here x is the first function, and cos x is the second function.
Using Integration by part
Evaluate the following integrals:
Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here x is the first function and is the second function.
Using Integration by part
Evaluate the following integrals:
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here x is the first function, and Sin 3x is the second function.
Using Integration by part
Evaluate the following integrals:
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here x is the first function, and Cos 2x is the second function.
Using Integration by part
Evaluate the following integrals:
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here log 2x is the first function, and x is the second function.
Using Integration by part
Evaluate the following integrals:
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here x is the first function, and cosec2x is the second function.
Using Integration by part
Evaluate the following integrals:
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here x2 is the first function, and cos x is the second function.
Using Integration by part
Again applying by the part method in the second half, we get
Evaluate the following integrals:
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Using Integration by part
Writing Sin2x =
We have
Taking X as first function and Cos 2x as the second function.
Evaluate the following integrals:
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Using Integration by part
Writing tan2x = sec2x - 1
We have
Using x as the first function and Sec2x as the second function
Evaluate the following integrals:
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here x2 is the first function, and ex is the second function.
Using Integration by part
Evaluate the following integrals:
We know that Cos3x = 4Cos3x - 3Cosx
Cos3x =
Taking X2 as the first function and cos 3x and cos x as the second function and applying By part method.
Evaluate the following integrals:
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here x2 is the first function, and e3x is the second function.
Using Integration by part
Evaluate the following integrals:
We can write
We have
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here x2 is the first function, and Cos 2x is the second function.
Using Integration by part
Evaluate the following integrals:
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here log2x is the first function, and x3 is the second function.
Using Integration by part
Evaluate the following integrals:
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here log(x + 1) is first function and x is second function.
Adding and subtracting 1 in the numerator,
Evaluate the following integrals:
We can write it as
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here logx is the first function, and x - n is the second function.
Evaluate the following integrals:
We can write it as
Let x2 = t
2xdx = dt
Using the relation in the above condition, we get
Integrating with respect to t
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here t is the first function, and et is the second function.
Replacing t with x2,we get
Evaluate the following integrals:
We know that Sin3x = 3Sinx - 4Sin3x
Sin3x = (3Sinx - Sin3x)/4
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here x is first function and sinx and sin3x as the second function.
Evaluate the following integrals:
We can write cos3x = (cos3x + 3cosx)/4, we have
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here x is first function and cosx and cos3x as the second function.
Evaluate the following integrals:
We can write it as
Now let x2 = t
2xdx = dt
Xdx = dt/2
Now
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here t is the first function and cost as the second function.
Replacing t with x2
=
Evaluate the following integrals:
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here log(cosx) is the first function and sinx as the second function.
Evaluate the following integrals:
We know that Sin2x = 2Sinxcosx
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here x is first function and sin2x as the second function.
Evaluate the following integrals:
Let √x = t
We can write it as
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here t is first function and cos t as the second function.
Replacing t with √x
= 2√xsin√x + 2cos√x + c
= 2(cos√x + √xsin√x) + c
Evaluate the following integrals:
We can write it as
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here cosecx is first function and cosec2x as the second function.
We know that Cot2x = Cosec2x - 1
We can write
Evaluate the following integrals:
We can write it as
We also know that 2sinx.cosx = sin2x
We also know that
Here Sin4x = 2sin2x.cos2x
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here x is first function and Sin2x and sin4x as the second function.
Evaluate the following integrals:
Let cosx = t
- sinxdx = dt
Now the integral we have is
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here logt is first function and 1 as the second function.
Replacing t with cosx
Evaluate the following integrals:
Let logx = t
1/x dx = dt
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here logt is first function and 1 as the second function.
Now replacing t with logx
Evaluate the following integrals:
=
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here log(2 + x2) is the first function and 1 as the second function.
Evaluate the following integrals:
We can write it as
Using by part and ILATE
Taking x as first function and sec2x and secxtanx as the second function, we have
Evaluate the following integrals:
Let us assume logx = t
X = et
dx = etdt
Now we have
Considering f(x) = 1/t ; f’(x) = - 1/t2
By the integral property of
So the solution of the integral is
Substituting the value of t as logx
Evaluate the following integrals:
We know that
Putting in the original equation
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here cos6x and cos2x is first function and e - x as the second function.
Solving both parts individually
Solving the second part,
Putting in the obtained equation
Evaluate the following integrals:
Let √x = t
Replacing in the original equation , we get
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here t is the first function and et as the second function.
Replacing t with √x
= 2e√x(√x - 1) + c
Evaluate the following integrals:
We can write Sin2x = 2sinx.cosx
Let Sinx = t
Cosxdx = dt
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here t is the first function and et as the second function.
Replacing t with sin x
= 2esinx(sinx - 1) + c
Evaluate the following integrals:
Let sin - 1x = t
X = sint
Putting this in the original equation, we get
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here t is the first function and sin t as the second function.
We can write cos t = √1 - sin2t
= - t(√1 - sin2t) + sint + c
Now replacing sin - 1x = t
= - sin - 1x(√1 - x2) + x + c
Evaluate the following integrals:
Let tan - 1 x = t and x = tan t
Differentiating both sides, we get
Now we have
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here t is the first function and sec2t as the second function.
We know that sec t = √tan2t + 1
Evaluate the following integrals:
We can write it as
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here log(x + 2) is first function and (x + 2) - 2 as second function.
Evaluate the following integrals:
Let x = sin t ; t = sin - 1x
dx = cos t dt
We know that sin 2t = 2 sint×cost
We have
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here t is the first function and sin 2t as the second function.
We know that cos2t = 1 - 2sin2t , sin2t = 2sint×cost and cos t = √1 - sin2t
Replacing in above equation
Evaluate the following integrals:
Let x = cos t ; t = cos - 1x
dx = - sin t dt
We know that sin 2t = 2 sint×cost
We have
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking first function to the one which comes first in the list.
Here t is first function and sin 2t as second function.
We know that cos2t = 2cos2t - 1 and sin2t = 2sint×cost and sint = √1 - cos2t
Replacing in above equation
Evaluate the following integrals:
We can write it as
Using BY PART METHOD. Using the superiority list as ILATE (Inverse Logarithm Algebra Trigonometric Exponential). Taking the first function to the one which comes first in the list.
Here cot - 1x is first function and 1 as the second function.
Let 1 + x2 = t
2xdx = dt
Xdx = dt/2
Now replacing t with 1 + x2
= xcot - 1x + log(1 + x2)/2 + c
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions , then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = cot-1x and f2(x) = x,
, where c is the integrating constant
Evaluate the following integrals:
[CBSE 2006C]
Tip – If f1(x) and f2(x) are two functions , then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = cot-1x and f2(x) = x2,
Taking (1+x2)=a,
2xdx=da i.e. xdx=da/2
Again, x2=a-1
Replacing the value of a, we get,
The total integration yields as
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions , then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = sin-1√x and f2(x) = 1,
Taking (1-x)=a2,
-dx=2ada i.e. dx=-2ada
Again, x=1-a2
Replacing the value of a, we get,
The total integration yields as
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions , then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = cos-1√x and f2(x) = 1,
Taking (1-x)=a2,
-dx=2ada i.e. dx=-2ada
Again, x=1-a2
Replacing the value of a, we get,
The total integration yields as
, where c is the integrating constant
Evaluate the following integrals:
Formula to be used – We know , cos3x = 4cos3x-3cosx
Assuming x = cosa, 4cos3a-3cosa=cos3a
And, dx = -sinada
Hence, a=cos-1x
Again, sina=√(1-x2)
Tip – If f1(x) and f2(x) are two functions , then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = a and f2(x) = sina,
Replacing the value of a we get,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions , then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = and f2(x) = 1,
Now,
Again, we know,
Replacing x by tanx, it is obtained that,
So, the final integral yielded is
, where c is the integrating constant
Evaluate the following integrals:
Formula to be used – We know,
Assuming x = tana,
And, dx = sec2ada
Hence, a=tan-1x
Now, sec2a-tan2a=1 , so,seca=√(1+x2)
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = a and f2(x) = sec2a,
Replacing the value of a we get,
, where c is the integrating constant
Evaluate the following integrals:
Formula to be used – We know,
Assuming x = tana,
And, dx = sec2ada
Hence, a=tan-1x
Now, sec2a-tan2a=1 , so, seca=√(1+x2)
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = a and f2(x) = sec2a,
Replacing the value of a we get,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions , then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = sin-1x and f2(x) = 1/x2,
Taking x= sina, dx = cosada
Hence, coseca=1/x
Now, cosec2a-cot2a = 1 so cota=√(1-x2)/x
Replacing the value of a, we get,
The total integration yields as
, where c is the integrating constant
Evaluate the following integrals:
Say, tanx = a
Hence, sec2xdx=da
Now, taking 1-a2 = k , -2ada=dk i.e. ada=-dk/2
Replacing the value of k,
Replacing the value of a,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions , then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = sin4x and f2(x) = e3x,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = sinx and f2(x) = e2x,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = sin2x and f2(x) = e2x,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = cos(3x+4) and f2(x) = e2x,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = cosx and f2(x) = e-x,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = sinx and f2(x) = ex in the first integral and keeping the second integral intact,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = cotx and f2(x) = ex in the first integral and keeping the second integral intact,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = secx and f2(x) = ex in the first integral and keeping the second integral intact,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = tan-1x and f2(x) = ex in the first integral and keeping the second integral intact,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = logsinx and f2(x) = ex in the second integral and keeping the first integral intact,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = logcosx and f2(x) = ex in the second integral and keeping the first integral intact,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = logcosx and f2(x) = ex in the second integral and keeping the first integral intact,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = tanx and f2(x) = ex in the second integral and keeping the first integral intact,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = cotx and f2(x) = ex in the first integral and keeping the second integral intact,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = secx and f2(x) = ex in the first integral and keeping the second integral intact,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = cotx and f2(x) = ex in the second integral and keeping the first integral intact,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = tan(x/2) and f2(x) = ex in the second integral and keeping the first integral intact,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = cot2x and f2(x) = ex in the first integral and keeping the second integral intact,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = sin-1x and f2(x) = ex in the first integral and keeping the second integral intact,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = logx and f2(x) = ex in the second integral and keeping the first integral intact,
, where c is the integrating constant
Evaluate the following integrals:
For x=-1, equation: -1 = B i.e. B = -1
For x=0, equation: 0 = A-1 i.e. A = 1
The given equation becomes
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = 1/(1+x) and f2(x) = ex in the first integral and keeping the second integral intact,
, where c is the integrating constant
Evaluate the following integrals:
For x=-1, equation: -2 = C i.e. C = -2
For x=0, equation: -1 = A+B-2 i.e. A+B = 1
For x=1, equation: 0 = 4A+2B-2
i.e. 2(A+B+A) = 2
⇨1+A = 1
⇨A = 0
And, B = 1
The given equation becomes
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = 1/(1+x)2 and f2(x) = ex in the first integral and keeping the second integral intact,
, where c is the integrating constant
Evaluate the following integrals:
For x=1, equation: 1 = B i.e. B = 1
For x=2, equation: 0 = -A+1 i.e. A = 1
The given equation becomes
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = 1/(1-x) and f2(x) = ex in the second integral and keeping the first integral intact,
, where c is the integrating constant
Evaluate the following integrals:
For x=1, equation: -2 = C i.e. C = -2
For x=0, equation: -3 = A-B-2 i.e. B = A+1
For x=3, equation: 0 = 4A+2B-2
i.e. 2(A+B+A) = 2
⇨1+3A = 1
⇨A = 0
And, B = 1
The given equation becomes
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = 1/(1-x)2 and f2(x) = ex in the first integral and keeping the second integral intact,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = 1/3xand f2(x) = e3x in the first integral and keeping the second integral intact,
, where c is the integrating constant
Evaluate the following integrals:
For x=-2, equation: -1 = B i.e. B = -1
For x=-1, equation: 0 = A-1 i.e. A = 1
The given equation becomes
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = 1/(x+2) and f2(x) = ex in the second integral and keeping the first integral intact,
, where c is the integrating constant
Evaluate the following integrals:
For x=-1/2, equation: -1/2 = B i.e. B = -1/2
For x=0, equation: 0 = A-1/2 i.e. A = 1/2
The given equation becomes
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = 1/(1+2x) and f2(x) = e2x in the second integral and keeping the first integral intact,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = 1/2x and f2(x) = e2x in the first integral and keeping the second integral intact,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = logx and f2(x) = ex in the first integral and keeping the second integral intact,
, where c is the integrating constant
Evaluate the following integrals:
For x=1, equation: 0 = A+B
For x=1/e, equation: -1 = B i.e. B = -1
So, A = 1
The given equation becomes
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = 1/(1+logx) and f2(x) = 1in the second integral and keeping the first integral intact,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = sin(logx) and f2(x) = 1in the first integral and keeping the second integral intact,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = 1/(logx) and f2(x) = 1in the first integral and keeping the second integral intact,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = log(logx) and f2(x) = 1in the first integral and keeping the second integral intact,
, where c is the integrating constant
Evaluate the following integrals:
It is know that sin-1x+cos-1x = π/2
Tip – If f1(x) and f2(x) are two functions , then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Now, for the first term,
Taking f1(x) = sin-1√x and f2(x) = 1,
Taking (1-x)=a2,
-dx=2ada i.e. dx=-2ada
Again, x=1-a2
Replacing the value of a, we get,
The total integration yields as
, where c’ is the integrating constant
For the second term,
Taking f1(x) = cos-1√x and f2(x) = 1,
Taking (1-x)=a2,
-dx=2ada i.e. dx=-2ada
Again, x=1-a2
Replacing the value of a, we get,
The total integration yields as
, where c’’ is the integrating constant
where c is the integrating constant
Evaluate the following integrals:
Tip – 5x is to be replaced by a
The equation becomes as follows:
Tip – 5a is to be replaced by k
The equation becomes as follows:
Re-replacing the value of k,
Re-replacing the value of a,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = tanx and f2(x) = e2x in the second integral and keeping the first integral intact,
, where c is the integrating constant
Evaluate the following integrals:
Tip – If f1(x) and f2(x) are two functions, then an integral of the form can be INTEGRATED BY PARTS as
where f1(x) and f2(x) are the first and second functions respectively.
Taking f1(x) = tanx and f2(x) = e2x in the second integral and keeping the first integral intact,
, where c is the integrating constant
Mark (√) against the correct answer in each of the following:
A. ex (1 – x) + C
B. ex (x – 1) + C
C. ex (x – 1) + C
D. none of these
To find: Value of
Formula used:
We have, … (i)
Ans ) c
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
To find: Value of
Formula used:
We have, … (i)
Ans ) B
Mark (√) against the correct answer in each of the following:
A.
B.
C. 2x sin 2x + 4 cos 2x + C
D. none of these
To find: Value of
Formula used:
We have, … (i)
Let 2x = t
Taking 1st function as and second function as
Mark (√) against the correct answer in each of the following:
A. x tan x – log |cos x| + C
B. x tan x + log |cos x| + C
C. x tan x + log |sec x| + C
D. none of these
To find: Value of
Formula used:
We have, … (i)
Taking 1st function as and second function as
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
To find: Value of
Formula used:
We have, … (i)
Let 2x = t
Taking 1st function as and second function as
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
To find: Value of
Formula used:
We have, … (i)
Taking 1st function as and second function as
Mark (√) against the correct answer in each of the following:
A. x cot x – log |sin x| + C
B. - cot x + log |sin x| + C
C. x tan x – log |sec x| + C
D. none of these
To find: Value of
Formula used:
We have, … (i)
Ans ) D None of these
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D.
To find: Value of
Formula used:
We have, … (i)
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
To find: Value of
Formula used:
We have, … (i)
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
To find: Value of
Formula used:
We have, … (i)
Mark (√) against the correct answer in each of the following:
A.
B.
C. x (log x + 1) + C
D. x (log x – 1) + C
To find: Value of
Formula used:
We have, … (i)
Taking 1st function as and second function as
Mark (√) against the correct answer in each of the following:
A.
B.
C. x (log x – 1) loge 10 + C
D. x(log x – 1) log10 e + C
To find: Value of
Formula used:
We have, … (i)
Taking 1st function as and second function as
Mark (√) against the correct answer in each of the following:
A.
B.
C. x (log x)2 – 2x log x + 2x + C
D. x (log x)2 + 2x log x – 2x + C
To find: Value of
Formula used:
We have, … (i)
Taking 1st function as and second function as
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
To find: Value of
Formula used:
We have, … (i)
Putting
⇒ dx = 2t dt
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
To find: Value of
Formula used:
We have, … (i)
Putting
⇒ dx = 2t dt
Mark (√) against the correct answer in each of the following:
A.
B.
C. 2x [cos (log x) + sin (log x)] + C
D. 2x [cos (log x) – sin (log x)] + C
To find: Value of
Formula used:
We have, … (i)
Taking cos(logx) as first function and 1 as second function.
Mark (√) against the correct answer in each of the following:
A.
B.
C. 2{sec x tan x + log |sec x + tan x|}+C
D. none of these
To find: Value of
Formula used:
We have, … (i)
Taking secx as first function and sec2x as second function.
Mark (√) against the correct answer in each of the following:
A. x log x + C
B.
C.
D. none of these
To find: Value of
Formula used:
We have, … (i)
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
To find: Value of
Formula used:
We have, … (i)
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
To find: Value of
Formula used:
We have, … (i)
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
To find: Value of
Formula used:
We have, … (i)
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
To find: Value of
Formula used:
We have, … (i)
Mark (√) against the correct answer in each of the following:
A. (2 sin x) esin x + C
B. (2 cos x) esin x + C
C. 2esin x (sin x + 1) + C
D. 2esin x (sin x – 1) + C
To find: Value of
Formula used:
We have, … (i)
Put sinx = t
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
To find: Value of
Formula used:
We have, … (i)
Putting sin-1x = t , x = sint
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
To find: Value of
Formula used:
We have, … (i)
Putting tan-1x = t , x = tant
dx = sec2t dt
When x = tant
Taking 1st function as and second function as
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
To find: Value of
Formula used:
We have, … (i)
Taking 1st function as and second function as
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
To find: Value of
Formula used:
We have, … (i)
Let,
⇒ dx = 2t dt
Taking 1st function as and second function as
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
To find: Value of
Formula used:
We have, … (i)
Let, ⇒ x = cosθ
⇒ dx = -sinθ dθ
If x = cosθ ,
Then = sinθ
Taking 1st function as and second function as
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
To find: Value of
Formula used:
We have, … (i)
Let, ⇒ x = tanθ
⇒ dx = sec2θ dθ
If x = tanθ ,
Then 1 + x2 = sec2θ
⇒ θ = sec-1
Taking 1st function as and second function as
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
To find: Value of
Formula used:
We have, … (i)
Let, ⇒ x = secθ
⇒ dx = secθ tanθ dθ
If x = secθ ,
Then = tanθ
Taking 1st function as and second function as
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
To find: Value of
Formula used:
We have, … (i)
Let x = sinθ , ⇒ θ = sin-1x
⇒ dx = cosθ dθ
If x = sinθ ,
Then = cosθ
Taking 1st function as and second function as
Mark (√) against the correct answer in each of the following:
A. 2x tan-1 x + log |1 + x2| + C
B. 2x tan-1 x - log |1 + x2| + C
C. 2x sin-1 x + log |1 + x2| + C
D. none of these
To find: Value of
Formula used:
We have, … (i)
Let x = tanθ , ⇒ θ = tan-1x
⇒ dx = sec2θ dθ
If x = tanθ ,
Then 1 + x2 = sec2θ
⇒ θ = sec-1
Taking 1st function as and second function as
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
To find: Value of
Formula used:
We have, … (i)
Let x = cosθ , ⇒ θ = cos-1x
⇒ dx = -sinθ dθ
If x = cosθ ,
Then = sinθ
Taking 1st function as and second function as
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D.
To find: Value of
Formula used:
We have, … (i)
Let x = tanθ , ⇒ θ = tan-1x
⇒ dx = sec2θ dθ
If x = tanθ ,
Then 1 + x2 = sec2θ
⇒ θ = sec-1
Taking 1st function as and second function as
Mark (√) against the correct answer in each of the following:
A. x2 sin x + 2x cos x – 2 sin x + C
B. 2x cos x – x sin x + 2 sin x + C
C. x2 sin x – 2x sin x + 2 sin x + C
D. none of these
To find: Value of
Formula used:
We have, … (i)
Taking 1st function as and second function as
Taking 1st function as and second function as
Mark (√) against the correct answer in each of the following:
A. cos x log (cos x) – cos x + C
B. -cos x log (cos x) + cos x + C
C. cos x log (cos x) + cos x + C
D. none of these
To find: Value of
Formula used:
We have, … (i)
Let cosx = t
-sinx dx = dt
Taking 1st function as and second function as 1
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
To find: Value of
Formula used:
We have, … (i)
Let 2x = t
2dx = dt
Taking 1st function as and second function as sint
Mark (√) against the correct answer in each of the following:
A. x2 sin x2 + cos x2 + C
B.
C.
D. none of these
To find: Value of
Formula used:
We have, … (i)
Let x2 = t
⇒ xdx = dt
Taking 1st function as and second function as
Mark (√) against the correct answer in each of the following:
A. 2x tan-1 x + log(1 + x2) + C
B. -2x tan-1 x – 2 log (1 + x2) + C
C. 2x tan-1 x – log (1 + x2) + C
D. none of these
To find: Value of
Formula used:
We have, … (i)
Let x = tant , t = tan-1x
⇒ dx = sec2t dt
If tant = x ,
sec t = 1 + x 2
Taking 1st function as and second function as
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
To find: Value of
Formula used:
We have, … (i)
Taking 1st function as and second function as
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
To find: Value of
Formula used:
We have, … (i)
Taking 1st function as sin(logx) and second function as 1
Taking 1st function as cos(logx) and second function as 1
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D.
To find: Value of
Formula used:
We have, … (i)
Putting sint = x, ⇒ t =
⇒ dx = cost dt
When x = sint then
Taking 1st function as t2 and second function as cost
Taking 1st function as t and second function as sint
Mark (√) against the correct answer in each of the following:
A.
B. xex – ex + C
C.
D. none of these
To find: Value of
Formula used:
We have, … (i)
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. none of these
To find: Value of
Formula used:
We have, … (i)
Mark (√) against the correct answer in each of the following:
A.
B. ex sin-1 x + C
C.
D. none of these
To find: Value of
Formula used:
We have, … (i)
Mark (√) against the correct answer in each of the following:
A. ex log sec x + C
B. ex tan x + C
C. ex (log cos x) + C
D. none of these
To find: Value of
Formula used:
We have, … (i)
Mark (√) against the correct answer in each of the following:
A. ex log sec x + C
B. ex tan x + C
C. ex (log cos x) + C
D. none of these
To find: Value of
Formula used:
We have, … (i)
Mark (√) against the correct answer in each of the following:
A. ex log (sec x + tan x) + C
B. ex sec x + C
C. ex log tan x + C
D. none of these
To find: Value of
Formula used:
We have, … (i)
Mark (√) against the correct answer in each of the following:
A.
B. ex tan-1 x + C
C. -ex cot-1 x + C
D. none of these
To find: Value of
Formula used:
We have, … (i)
Mark (√) against the correct answer in each of the following:
A. ex tan x + C
B. ex log cos x + C
C. ex log sec x + C
D. none of these
To find: Value of
Formula used:
We have, … (i)
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. None of these
To find: Value of
Formula used:
We have, … (i)
Mark (√) against the correct answer in each of the following:
A. ex sin x + C
B. ex cos x + C
C. ex tan x + C
D. None of these
To find: Value of
Formula used:
We have, … (i)
Mark (√) against the correct answer in each of the following:
A. ex (1 + tan x) + C
B. ex sec x + C
C. ex tan x + C
D. none of these
To find: Value of
Formula used:
We have, … (i)
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. None of these
To find: Value of
Formula used:
We have, … (i)
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. None of these
To find: Value of
Formula used:
We have, … (i)
Mark (√) against the correct answer in each of the following:
A.
B.
C.
D. None of these
To find: Value of
Formula used:
We have, … (i)