Integrals

Method: To find the anti derivative of a function by inspection
Steps: 1. In this method we look for a function whose derivative is the given function. For Example: if we need to find anti derivative of x², we know that derivative of x³ is 3x². Therefore, the variable terms comes out to be the same.
2. After that balance out the coefficients of variables by dividing and multiplying suitable terms. From above example if [we divide x³ by 3 we will get the answer as x². Hence, we can say that anti derivative of x² is x³/3.

Now, similarly,

We know that

Therefore, the anti-derivative of sin2x is

We know that

Therefore, the anti-derivative of cos3x is .

We know that

Therefore, the anti-derivative of is .

We know that

Therefore, the anti-derivative of is .

We know that

Therefore, the anti-derivative of sin2x is …(1)

Also,

Therefore, the anti-derivative of is …(2)

From (1) and (2), we get,

= sin 2x – 4e^3x

Therefore, the anti-derivative of sin 2x – 4 e^3x is .

Separating the terms we get,

Separating the terms we get,

∫ xⁿ dx =

Now the numerator can be factorized as,

x³ - x² + x - 1 = x²(x - 1) + 1(x - 1)

x³ - x² + x - 1 = (x² + 1)(x - 1)

Now putting this in given integral we get,

Formulas Used: ∫ sec²x dx = tanx + c and ∫ secx tanx dx = sec x + c

Opening the brackets we get,

= tanx –x +C

= 2tanx – 3secx + C

It is given that

Also, It is given that f(2) = 0

Therefore,

Given:

Let x² + 1 = t

⇒ 2xdx = dt

⇒ xdx = � dt

When x = 0, t = 1 and when x = 1, t = 2

Given:

Let

Also, let

when,

so,

Given:

Let x = tan θ ⇒ dx = sec² θ d θ

When, x = 0, θ = 0 and when x = 1, θ = π /4

Let

By applying product rule as,

Given:

Let x + 2 = t² ⇒ dx = 2t dt

And x = t² -2

when, x = 0, t = √2 and when x = 2, t = 2

so,

Given:

Let cos x = t

⇒ -sin xdx = dt

⇒ sin xdx = -dt

When x = 0, t = 1 and when x = π /2, t = 0

because,

Given:

we can write it as,

let

when

because,

Given:

Let x + 1 = t

⇒ dx = dt

When x = -1, t = 0 and when x = 1, t = 2

because,

Given:

Let 2x = t ⇒ 2 dx = dt

When x = 1, t = 2 and when x = 2, t = 4

now, let 1/t = f(t)

then, f' (t) = -1/t²

because,

Given:

let

Now, let x = sin θ ⇒ dx = cos θ d θ

when, and when x = 1, θ = π/2

Now, let cot θ = t ⇒ - cosec² θ d θ

when,

= 6

Correct option is: (A)

Given:

Applying product rule,

⇒

= -x cos x + sin x – 0

⇒ f(x) = -x cos x + sinx

⇒ f'(x) = -[{x(-sinx )} + cosx ] + cosx

= x sin x – cos x + cos x

= x sin x

Correct answer is B.

Given:

let,

as,

Adding (1) and (2), we get

Given:

let,

as,

Adding (1) and (2), we get

Given:

Adding (1) and (2), we get

Given:

Adding (1) and (2), we get

Given:

As we can see that (x+2) ≤ 0 on [-5, -2] and (x+2) ≥ 0 on [-2,5]

Given:

As we can see that (x-5)≤0 on [2,5] and (x+2)≥0 on [5,8]

⇒ I = 9

Given:

Given:

Given:

Given:

Adding (1) and (2), we get

Given:

As we can see f(x) = sin²x and f(-x) = sin²(-x) = (sin (-x))² = (-sin x)² = sin²x.

i.e. f(x) = f(-x)

so, sin²x is an even function.

It is also known that if f(x) is an even function then,

Given:

Adding (1) and (2), we get

⇒ I = π

Given:

As we can see f(x) = sin⁷x and f(-x) = sin⁷(-x) = (sin (-x))⁷ = (-sin x)⁷ = -sin⁷x.

i.e. f(x) = -f(-x)

so, sin²x is an odd function.

It is also known that if f(x) is an odd function then,

Given:

As we see, f(x)=cos⁵x and f(2π –x) = cos⁵(2π –x) = cos⁵x = f(x)

Given:

Adding (1) and (2), we get

Given:

Adding (1) and (2), we get

Here, if f(x) = log (sin x) and f(π – x)=log ( sin (π –x))= log (sin x ) = f(x)

Adding (1) and (2), we get

Let 2x = t ⇒ 2dx = dt

When x = 0, t = 0 and when x = π /2, t = π

Given:

Adding (1) and (2), we get

Given:

As we can see that (x-1)≤0 when 0≤x≤1 and (x-1)≥0 when 1≤x≤4

Given:

Adding (1) and (2), we get

Given:

It is also known that if f(x) is an even function then,

It is also known that if f(x) is an odd function then,

Correct answer is C

Given:

Adding (1) and (2), we get

Correct answer is (c)

Given:

Let

Using partial differentiation:

let

⇒ 1 = A – Ax² + Bx – Bx² + Cx + Cx²

⇒ 1 = A + (B+C)x + (-A-B+C)x²

Equating the coefficients of x, x² and constant value. We get:

(a) A = 1

(b) B+C = 0 ⇒ B = -C

(c) -A –B +C =0

⇒ -1 – (-C) +C = 0

⇒ 2C = 1 ⇒ C = 1/2

So, B = -1/2

Put these values in equation (1)

Given:

let

Multiply and divide by,

Given:

let

put

Given:

let

Multiply and divide by x^-3, we get

Given:

or we can write it as,

Let x = t⁶⇒ dx = 6t⁵dt

After division we get,

Given:

Using partial differentiation:

let

⇒5x=A(x²+9)+(Bx+C)(x+1)

⇒ 5x = Ax² +9A+ Bx² +Bx+ Cx + C

⇒ 5x = 9A + C + (B+C)x + (A+B)x²

Equating the coefficients of x, x² and constant value. We get:

(a) 9A + C = 0 ⇒ C = -9A

(b) B+C = 5 ⇒ B = 5-C ⇒ B = 5-(-9A) ⇒ B = 5 + 9A

( c) A + B =0 ⇒ A = -B ⇒ A = -(5 + 9A) ⇒ 10A = -5 ⇒ A = -1/2

and C = 9/2 and B = 1/2

Put these values in equation (1)

Put x² = t ⇒ 2xdx = dt

Put the value in equ. (2)

Given:

Let x – a = t ⇒ x = t + a ⇒ dx = dt

As,{ sin (A+B) = sin A cos B + cos A sin B}

Given:

Given:

Put sin x = t ⇒ cos x dx = dt

Given:

Given:

Multiply and divide by sin (a-b), we get

As,{ sin (A-B) = sin A cos B - cos A sin B}

Given:

Now, let x⁴ = t ⇒ 4x³ dx = dt

And x³ dx = dt/4

Given:

Let e^x = t ⇒ e^x dx = dt

Given:

Using partial differentiation:

⇒1 = (Ax + B)(x² + 4)+(Cx + D)(x² + 1)

⇒ 1 = Ax³ +4Ax+ Bx² + 4B+ Cx³ + Cx + Dx² + D

⇒ 1 = (A+C)x³ +(B+D)x² +(4A+C)x + (4B+D)

Equating the coefficients of x, x², x³ and constant value. We get:

(a) A + C = 0 ⇒ C = -A

(b) B + D = 0 ⇒ B = -D

( c) 4A + C =0 ⇒ 4A = -C ⇒ 4A = A ⇒ 3A = 0 ⇒ A = 0 ⇒ C = 0

( d) 4B + D = 1 ⇒ 4B – B = 1 ⇒ B = 1/3 ⇒ D = -1/3

Put these values in equation (1)

Given:

Let

Let cos x = t ⇒ -sin x dx = dt ⇒ sin x dx = dt

Given:

Let

Let x⁴ = t ⇒ 4x³ dx = dt ⇒ x³ dx = dt/4

Given: f^’ (ax + b)[f(ax + b)]ⁿ

Let f(ax +b) = t ⇒ a .f^’ (ax + b)dx = dt

Given:

As,{ sin (A+B) = sin A cos B + cos A sin B}

Given:

let

As we know,

Now, first solve for I₁:

because,

Put it in equ. (2)

Given:

let

Let x= cos²θ ⇒ dx = -2sinθ cosθ dθ

Given:

Now let tan x = f(x)

⇒ f^’(x) = sec²x dx

Given:

Let

Using partial differentiation:

let

⇒ x² + x + 1 = Ax² + 3Ax + 2A + Bx +2B + Cx² + 2Cx + C

⇒ x² + x + 1 = (2A+2B+C) + (3A+B+2C)x + (A+C)x²

Equating the coefficients of x, x² and constant value. We get:

(a) A + C = 1

(b) 3A + B + 2C = 1

( c) 2A+2B+C =1

After solving we get:

A=-2, B=1 and C=3

Given:

Let x= cosθ ⇒ dx = -sin θ d θ

Given:

Given:

Given:

Now let tan²x = t ⇒ 2 tan x sec²x dx = dt

And when x=0 then t=0 and when x=π /4 then t=1

Given:

First solve for I1:

Let 2 tan x = t ⇒ 2 sec²x dx dt

When x = 0 then t = 0 and when x = π /2 then t = ∞

Put this value in equ.(2)

Given:

Now let sin x – cos x = t ⇒( cos x + sin x )dx = dt

i.e. f(x) = f(-x)

so, f(x) is an even function.

It is also known that if f(x) is an even function then,

Given:

Let

Also, let sinx – cosx = t

Differentiating both sides, we get,

(cosx + sinx) dx = dt

When x = 0, t = -1

And when , t = 0

Now,

(sinx – cosx)² = t²

1 – 2 sinx.cosx = t²

Sin2x = 1 – t²

Putting all the values, we get the integral,

Given:

Let sin x = t ⇒ cos x dx = dt

When x =0 then t = 0 and when x = π /2 then t = 1

Given:

Adding (1) and (2), we get

Given:

First solve for I1:

As we can see that (x-1)≥0 when 1≤x≤4

Now solve for I2:

As we can see that (x-2)≤0 when 1≤x≤2 and (x-2)≥0 when 2≤x≤4

Now solve for I3:

As we can see that (x-3)≤0 when 1≤x≤3 and (x-3)≥0 when 3≤x≤4

Given:

Using partial differentiation:

⇒ 1 = Ax² +Ax+ B+Bx+ Cx²

⇒ 1 = B + (A+B)x + (A+C)x²

Equating the coefficients of x, x² and constant value. We get:

(a) B = 1

(b) A + B = 0 ⇒ A = -B ⇒ A = -1

( c) A + C =0 ⇒ C = -A ⇒ C = 1

Put these values in equation (1)

⇒ L.H.S = R.H.S

Hence proved.

Given:

L.H.S = R.H.S

Hence Proved.

Given:

As we can see f(x) =x¹⁷ .cos⁴x and f(-x) = (-x)¹⁷ .cos⁴(-x) = -x¹⁷ .cos⁴x

i.e. f(x) = -f(-x)

so, it is an odd function.

It is also known that if f(x) is an odd function then,

Hence proved.

Given:

First solve for I1:

Let cos x = t ⇒ -sin x dx = dt ⇒ sinx dx = -dt

When x=0 then t= 1 and when x = π /2 then t = 0

Put in equ. (2)

L.H.S = R.H.S

Hence Proved.

Given:

First solve for I1:

Let tan x = t ⇒ sec² x dx = dt

When x=0 then t= 0 and when x = π /2 then t = 1

Put in equ. (2)

L.H.S = R.H.S

Hence Proved.

Given:

First solve for I1:

Let 1 - x² = t ⇒ -2 x dx = dt

When x=0 then t= 1 and when x = 1 then t = 0

Put in equ. (2)

L.H.S = R.H.S

Hence Proved.

Given:

Here, a=0, b=1, and f(x)=e^2-3x and h = 1/n

Given:

Put e^x= t ⇒ e^x dx = dt

Hence, correct option is (A).

Given:

Put sin x + cos x= t ⇒ cos x – sin x = dt

Hence, correct option is (B).

Given:

Hence, correct option is (D).

Given:

Adding (1) and (2), we get

Hence, correct option is (B).

Let 1+x² = t

⇒ 2x dx = dt

Now,

= log|t| + C

= log|1+x²| + C

= log(1+x²) + C

Let log|x| = t

Now,

⇒

⇒

let 1+logx =t

= log|t| + C

= log | 1+ logx| + C

Let cosx = t

⇒ -sinxdx = dt

⇒

=-[-cost] + C

= cost + C

= cos(cosx) + C

Let I = ∫sin (ax + b) cos (ax + b) dx

We know that,

sin 2A = 2sinA.cosA

Therefore,

sin (ax + b) cos (ax + b) =

Let 2(ax+b) = t

⇒ 2adx = dt

⇒

=

Let ax + b =t

⇒ adx = dt

Let (x +2) = t

⇒ dx = dt

Let 1 +2x² =t

⇒ 4xdx = dt

⇒

⇒

⇒

⇒

Let x² + x+ 1= t

Differentiating both sides, we get,

⇒ (2x + 1)dx = t dt

Therefore,

⇒

⇒

⇒

⇒

Now, Let

⇒

⇒

= 2log|t| + C

= 2log|| + C

Let x+ 4 = t

⇒ dx = dt

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

Let x³ -1 = t

⇒ 3x²dx = dt

⇒

⇒

⇒

⇒

⇒

⇒

Let 2 +3x³ = 1

⇒ 9x² dx = dt

⇒

⇒

⇒

⇒

Let log x = t

⇒

⇒

⇒

⇒

Let 9 – 4x² = t

⇒ -8xdx = dt

⇒

⇒

⇒

Let 2x +3 =t

⇒ 2dx = dt

⇒

⇒

⇒

Let x² = t

⇒ 2xdx = dt

⇒

⇒

⇒

⇒

⇒

let tan-1 x = t

⇒

⇒

= e^t + C

=

We have,

Dividing numerator and denominator by e^x, we get,

Let e^x + e^-x = t
Differentiating both sides, we get,

= log |t| +C

= log|e^x + e^-x| + C

Let

⇒

⇒

⇒

⇒

⇒

= log |t| +C

= log|| + C

tan² (2x – 3) = sec² (2x – 3) – 1

Let 2x -3 = t

⇒ 2dx = dt

⇒

⇒

⇒

⇒ 1/2 tan t - t + C

⇒ 1/2 tan (2x - 3) - (2x - 3) + C

Let 7 – 4x = t

⇒ -4dx = dt

⇒

⇒

⇒

let sin^-1 x =t

⇒

⇒

⇒

⇒

let 3cosx + 2sinx = t

(-3sinx + 2cosx)dx = dt

⇒

⇒

⇒

⇒

Let (1 – tanx) = t

⇒ -sec²xdx = dt

⇒

⇒

⇒

Let

⇒

⇒

= 2sint + C

= 2sin + C

Let sin2x = t

⇒ 2cos2xdx = dt

.

⇒

⇒

Let 1 + sinx = t

⇒ cosx dx = dt

⇒

⇒

⇒

⇒

Let Log sinx = t

⇒

⇒ cotx dx = dt

⇒

⇒

⇒

Let 1+ cosx = t

⇒ -sinx dx = dt

⇒

= - log|t| + C

= -log|1 + cosx| + C

Let 1 + cosx = t

⇒ -sinx dx = dt

⇒

⇒

⇒

⇒

Let I =

⇒

⇒

⇒

⇒

⇒

⇒

Let sinx + cosx = t

⇒ (cosx-sinx)dx = dt

Therefore, I =

⇒

⇒

Let I =

⇒

⇒

⇒

⇒

⇒

⇒

⇒

Let cosx - sinx = t

⇒ (-sinx-cosx)dx = dt

Therefore, I =

⇒

⇒

Let I =

⇒

⇒

⇒

Let tanx = t

⇒ sec²x dx = dt

⇒ I =

= 2+C

= 2 + C

let 1 + log x = t

⇒ = dt

⇒

⇒

⇒

Let (x+logx) = t

⇒

⇒

⇒

⇒

Let x⁴ = t

⇒ 4 x³ dx = dt

⇒

⇒

Let tan-1 = v

⇒

Thus, we get,

⇒

⇒

⇒

⇒

Let x¹⁰ + 10^x = t

⇒ (10x⁹ + 10^x log_e10)dx = dt

⇒

= log t + C

= log(x¹⁰ + 10^x) + C

Let I =

⇒

⇒

⇒

= tanx – cotx + C

sin²(2x+5)

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

Let I = sin³(2x+1)

⇒

⇒

Let cos (2x+1) = t

=> -2sin(2x+1)dx = dt

=> sin(2x+1)dx =

⇒

⇒

⇒

⇒

Let I =

⇒

⇒

Let cosx = t

⇒ -sinx.dx = dt

⇒ I

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

sin⁴x = sin²xsin²x

⇒

⇒

⇒

⇒

⇒

⇒

Now,

⇒

⇒

⇒

cos⁴2x = (cos²2)²

⇒

⇒

⇒

⇒

⇒

⇒

Now,

⇒

⇒

⇒

= 1- cosx

⇒

= x – sinx + C

Using the identity , we have

Now, using the identity sin 2x = 2 sin x cos x, we have

= 2[cos(x) +cosα]

= 2cosx + 2 cosα

⇒

= 2[sinx + xcosα] + C

⇒

⇒

Let sinx + cosx = t

⇒ (cosx-sinx)dx =dt

⇒

⇒

= -t^-1 + C

⇒

⇒

tan³2xsec2x = tan²2xtan2xsec2x

= (sec²2x -1)tan2xsec2x

= sec²2x.tan2xsec2x-tan2xsec2x

⇒

⇒

Now, Let sec2x = t

⇒ 2sec2xtan2x dx = dt

Thus,

⇒

⇒

tan⁴x = tan²x.tan²x

= (sec²x-1) tan²x

= sec²x tan²x- tan²x

= sec²x tan²x- (sec²x-1)

= sec²x tan²x- sec²x+1

Now,

⇒

Now, let tanx = t

=> sec²x dx =dt

⇒

⇒

⇒

⇒

= tanxsecx + cotxcosecx

⇒ Now,

= secx – cosecx + C

⇒

= sec²x

Now,

= tanx + C

⇒

⇒

Now,

let tanx = t

⇒ sec²x dx = dt

⇒

⇒

⇒

Now,

Let 1 + sin2x = t

⇒ 2cos2x dx = dt

Thus,

⇒

⇒

⇒

= log|sinx + cosx| + C

sin^-1(cosx)

Let cosx = t.......(1)

Then, sinx =

Differentiating both sides of (1), we get,

⇒ (-sinx)dx = dt

⇒ dx =

⇒ dx =

⇒ Now,

⇒

Let sin^-1 t = v

⇒

⇒ ∫sin^-1(cosx) dx = - ∫vdv

⇒

⇒

…(1)

We know that,

sin-1x + cos-1 x =

⇒ sin^-1(cosx)=

Now, substituting in eq(1), we get,

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

Now,

⇒

⇒

We know that,
∫sec²x dx = tan x + c
∫cosec²x dx = - cot x + c
= tanx + cotx + C.
A: Answer

Let x.e^x = t
Differentiating both sides we get,

⇒ (e^x.x + e^x.1)dx = dt

⇒ e^x(x + 1) = dt

Now,

= tan t + C

= tan(e^x.x) + C

Let x³ = t

⇒ 3x² dx = dt

⇒

= tan^-1t + C

= tan^-1(x³) + C

Let 2x= t

⇒ 2dx = dt

⇒

⇒

Let 2 – x = t

⇒ -dx = dt

⇒

⇒

⇒

Let 5x = t

⇒ 5dx = dt

⇒

⇒

⇒

Let = t

⇒ 2dx = dt

⇒

⇒

⇒

Let x³ = t

⇒ 3x² dx = dt

⇒

⇒

⇒

For ,

Let x² -1 = t

⇒ 2x dx = dt

⇒

⇒

⇒

⇒

⇒

Let x³ = t

⇒ 3x² dx = dt

⇒

⇒

⇒

Let tanx = t

⇒ sec²xdx = dt

⇒

⇒

⇒

Let x+1 = t

⇒ dx = dt

⇒

⇒

⇒

⇒

⇒

Let 3x+1= t

⇒ 3dx = dt

⇒

⇒

⇒

Let x+3= t

⇒ dx = dt

⇒

⇒

⇒

Let

⇒ dx = dt

⇒

⇒

⇒

Let

⇒ dx = dt

⇒

⇒

⇒

(x - 1)(x - b) can be written as x² – (a+b)x + ab.

Then, x² – (a+b)x + ab = x² – (a+b)x +

⇒

⇒

Let x =

⇒ dx = dt

⇒

⇒

⇒

Let 4x + 1 =

⇒ 4x + 1 = A(4x + 1) + B

⇒ 4x + 1 = 4Ax+ A+ B

Now, equating the coefficients of x and constant term on both sides, we get,

4A = 4

⇒ A = 1

A + B = 1

⇒ B = 0

Let 2x² + x – 3 =t

⇒ (4x + 1) dx = dt

⇒

⇒

2

Let x + 2 =

⇒ x + 2 = A(2x) + B

Now, equating the coefficients of x and constant term on both sides, we get,

2A = 1

⇒ A =

B = 2

⇒ x + 2 = (2x) + 2

⇒

⇒

Now,

Let x² - 1 = t

⇒ (2x)dx = dt

⇒

⇒

And

⇒

Let 5x - 2 =

⇒ 5x - 2 = A(2 + 6x) + B

Now, equating the coefficients of x and constant term on both sides, we get,

6A = 5

⇒ A =

2A + B = -2

⇒ B =

⇒ 5x - 2 = (2 + 6x) +

⇒

⇒

Now,

Let 1+2x+3x² = t

⇒ (2 + 6x)dx = dt

⇒

= log|1+2x+3x²| …(1)

And

1+2x+3x² =

⇒

⇒

⇒

⇒

…(2)

Thus, from (1) and (2), we get,

⇒

⇒

Let 6x + 7 =

⇒ 6x + 7 = A(2x - 9) + B

Now, equating the coefficients of x and constant term on both sides, we get,

2A = 6

⇒ A = 3

-9A + B = 7

⇒ B = 34

⇒ 6x + 7 = 3 (2x - 9) + 34

⇒

⇒

Now,

Let x² – 9x + 20 = t

⇒ (2x - 9)dx = dt

----------(1)

And

x² – 9x + 20 =

⇒

⇒

…(2)

Thus, from (1) and (2), we get,

⇒

⇒