Definite Integrals

Evaluation:

Evaluation:

Evaluation:

Evaluation:

Evaluation:

=[log(-1)-log(-4)]

=-[log(-4)-log(-1)]

=-log 4

Evaluation:

=[2√4-2]

=[4-2]

=2

Evaluation:

Evaluation:

=[6-3]

=3

6

Evaluation:

=3[4-2]

=6

Evaluation:

=[tan^-1 1-tan^-1 0]

=π/4

_Evaluation:

=[tan^-1 ∞-tan^-1 0]

=π/2

_Evaluation:

=[sin^-1 1-sin^-1 0]

_Evaluation:

-2

_Evaluation:

=-2

_Evaluation:

_Evaluation:

Evaluation:

Evaluation:

log 2

Evaluation:

=log|2|-log|1|

=log2

Evaluation:

=-log|√2+1|+log|2+√3|

_Evaluation:

_Evaluation:

_Evaluation:

1

Evaluation:

=1

_Evaluation:

=[sin x + cos x]

=[√2-1]

2

Evaluation:

Let

=2

_Evaluation:

_Evaluation:

_Evaluation:

_Evaluation:

_Evaluation:

2

Explanation:

=2

2

Explanation:

=2

_Explanation:

(2 log 3 – 3 log 2)

Explanation:

=2log3-3log2

_Explanation:

_Evaluation:

Evaluation:

Substitute:

Undo substitution:

_Evaluation:

Substitute:

x+2=u

∴ dx=du

Undo substitution:

=log(4+√15)-log(3+√8)

Evaluation:

Substitute 4x+1√7=u

Now solving:

_Evaluation:

_Evaluation:

Substitute:

tan(x)=u

_Evaluation:

_{By reduction formula:}

We know that,

_Evaluation:

Assume that a≠0.

Now,

Substitute:

u=2x+(-√5-1)a

Undo substitution:

Now,

Substitute:

Undo substitution:

_Evaluation:

Substitute:

2x-1=u

Undo Substitution:

u=2x-1

∴=sin^-1 (2x-1)

_Evaluation:

Substitute:

2x-1=u

Substitute:

u=sin(v)

∴sin^-1 (u)=v

∴du=cos(v)dv

We know that,

Undo Substitution:

v=sin^-1 (u)
sin(sin^-1 (u))=u

Undo Substitution:

u=2x-1

_Evaluation:

Perform partial fraction decomposition:

_Evaluation:

1

Evaluation:

=[(x-1)e^x ]

=[(1-1) e¹-(0-1) e⁰]

=1

Evaluation:

Evaluation:

From integrate by parts:

From integrate by parts:

_Evaluation:

_Evaluation:

_Evaluation:

(2 log 2 – 1)

Evaluation:

_Evaluation:

Now,

Let,

∴dx=-x² du

Undo substitution:

_{Correct answer is}

Evaluation:

Let,

log(x)=u

→x=e^u

→dx=e^u du

Undo substitution:

Evaluation:

=log(x) e^x

=log(e) e^e-log(1) e¹

=e^e

_Evaluation:

From Integrates by parts:

=-e^x

(1 – log 2)

Evaluation:

Substitute:

Undo substitution:

=1-log2

_Explanation:

Let

Let 2x-3=t

⇒ 2dx=dt.

Hence,

Let

Let 1+x²=t

⇒ 2xdx=dt.

Also,

when x=0, t=1

and

when x=1, t=2

Hence,

Let

Let 9x²-1=t

⇒ 18xdx=dt.

Also,

when x=1, t=8

and

when x=2, t=35.

Hence,

Let

Let tan^-1x=t

⇒ .

Also, when x=0, t=0

and when x=1,

Hence,

Let

Let e^x=t

⇒ e^x dx=dt.

Also,

when x=0, t=1

and

when x=1, t=e.

Hence,

Let

Let x²=t

⇒ 2xdx=dt.

Also,

when x=0, t=0

and

when x=1, t=1.

Hence,

Let

Let x²=t

⇒ 2xdx=dt.

Also,

when x=0, t=0

and

when x=1, t=1.

Hence,

Let

Let

⇒ .

Also,

when x=1, t=1

and

when x=2, .

Hence,

Let

Let 3+4sinx=t

⇒ 4cosxdx=dt.

Also,

when x=0, t=3

and

when , t=5.

Hence,

Let

Let cos x=t

⇒ -sin x dx=dt.

Also,

when x=0, t=1

and

when , t=0.

Hence,

Let

Let e^x=t

⇒ e^x dx=dt.

Also,

when x=0, t=1

and

when x=1, t=e.

Hence,

Let

Let

⇒ .

Also,

when , t=-1

and

when x=e, t=1.

Hence,

=0

Let

Let tan^-1x=t

.

Also,

when x=0, t=0

and

when x=1,

Hence,

Let

Let 1+cos x=t

⇒ -sin x dx=dt.

Also, when x=0, t=2

and

when , t=1

Hence,

=2(√2-1)

Let

Let sinx=t

⇒ cos x dx=dt.

Also,

when x=0, t=0

and

when , t=1.

Consider cos⁵x=cos⁴x×cosx=(1-sin²x)²×cosx (Using sin²x+cos²x=1)

Hence,

Let

Let sin²x=t

⇒ 2sin x cos x=dt.

Also,

when x=0, t=0

and

when , t=1.

Hence,

Let

Let x=a sin t

⇒ a cos t dt=dx.

Also,

when x=0, t=0

and

when x=a, .

Hence,

Using , we get

Let

Consider,

Let x=a sin t

⇒ a cos t dt=dx.

Also, when x=0, t=0

and when x=a, .

Hence,

Using , we get

Here , hence

Let

Let x=a sin t

⇒ a cos t dt=dx.

Also, when x=0, t=0

and when x=a, .

Hence,

Using , we get

⇒

Hence,

Let

Let a²+x²=t²

⇒ x dx=t dt.

Also, when x=0, t=a

and when x=a, .

Hence,

=a(√2-1)

Let

Using the property that , we get

Hence,

Let

Let

Let x=tanθ

⇒ θ=tan^-1x

=sin^-1 (2sinθcosθ)

=sin^-1 (sin2θ)

Hence f(x)=2θ

=2tan^-1x

Hence

Using integration by parts, we get

-(1)

Let

Let 1+x²=t

⇒ 2x dx=dt.

Also, when x=0, t=1

and when x=1, t=2

Hence,

–(2)

Substituting value of (2) in (1), we get

Let

Using , we get

=2

Let

Using and

=-(√2-2) +(√2)

=2

Let

Dividing by cos²x in numerator and denominator, we get

Let tan x=t

⇒ sec²xdx=dt

Let

Let

Dividing by cos²x in numerator and denominator, we get

Consider

Let tan x=t

⇒ sec²xdx=dt

Let

=tan x

Here, a=1 and b=√2

Hence,

Let

Dividing by cos²x in numerator and denominator, we get

Consider

Let tan x=t

⇒ sec²xdx=dt

Let

=tan x

Here, a=2 and b=√13

Hence,

Let

Using , we get

Let

⇒ ,

when x=0, t=0 and when , t=1.

Hence,

Let

⇒ dt=du.

When t=0, and when t=1, .

Let

Using , we get

Let

,

when x=0, t=0 and when x=π, t=∞.

Hence,

Let

Using , we get

Let

⇒ ,

when x=0, t=0 and when x=π, t=∞.

Hence,

Let

Using

And

,

we get

Let

⇒ ,

when x=0, t=0

and when , t=1.

Hence,

Let t-2=u

⇒ dt=du.

Also, when t=0, u=-2

and when t=1, u=-1.

Hence,

(Using )

Let

Using

And

,

we get

Let

when x=0, t=0

and when , t=∞.

Hence,

Let t+1=u

⇒ dt=du.

Also, when t=0, u=1

and when t=∞, u=∞.

Let

Using 1+cos2x=2cos²x, we get

Let tan x=t

⇒ sec²xdx=dt.

when x=0, t=0

and when , t=1.

Let

Let cos x=t

⇒ -sin x dx=dt.

Also, when x=0, t=1

and when , t=0.

Hence,

Hence

Let

Using sin 2x =2 sin x cos x, we get

Let tan x=t

⇒ sec²xdx=dt.

Also, when x=0, t=0

and when , t=∞.

Hence,

Let x²=t

⇒ 2xdx=dt.

Also, when x=0, t=0

and when x=∞, t=∞.

Hence,

Let

Using

And

,

we get

Let

⇒ .

Also, when ,

and when , t=1

Hence,

Let

Let x=cost ⇒ dx=-sin t dt.

Also, when x=0,

and when x=1, t=0.

Hence,

Using integration by parts, we get

Hence, I=π-2

Let

Using integration by parts, we get

Let tan^-1x=t

⇒ .

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

Hence

Let 1+x²=y

⇒ 2xdx=dy.

Also, when x=0, y=1

and when x=1, y=2.

.

Let

Let √x=t

⇒

or

dx=2tdt.

When, x=0, t=0

and when x=1, t=1.

Hence,

Using integration by parts, we get

Let t=sin y

⇒ dt=cos y dy.

When t=0, y=0, when t=1, .

….. (1)

Using, , we get

…..(2)

Adding (1) and (2), we get

Hence,

Let

Let x=a tan²y

⇒ dx=2a tan y sec²y dy.

Also, when x=0, y=0

and when x=a,

Hence

Using integration by parts, we get

Let tan y=t

⇒ sec²ydy=dt.

Also, when y=0, t=0

and when , t=1.

Also, y=tan^-1t

Let

Hence

Substituting value of I’ in I, we get

Let

Let √x=u

or dx=2udu.

Also, when x=0, u=0 and x=9, u=3.

Hence,

Let

Let 1+3x⁴=t

⇒ 12x³dx=dt.

Also, when x=0, t=1 and when x=1, t=4.

Let

Let

Let x=tan t

⇒ dx=sec²tdt.

Also when x=0, t=0 and when x=1, .

Hence,

Using , we get

Let

Let 1+x²=t ⇒ 2xdx=dt.

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

Substituting t=1+x²

⇒ 2xdx=dt.

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

Hence,

Let

Let x=sect

⇒ dx=sec t tan t dt.

Also,

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

Hence,

Using , we get

Let

Let sin x- cos x=t

⇒ (cos x + sin x)dx=dt.

When x=0, t=-1 and , t=1.

Also, t²=(sin x – cos x)²

=sin²x+cos²x-2sinxcosx

=1-2sinxcosx

or

Hence

Let t=sin y

⇒ dt=cos y dy.

Also, when t=-1,

and when t=1, .

Let

Let,

=-2ax+5a+b

Hence -2a=-1 and 5a+b=2.

Solving these equations,

we get and .

We get,

Let

Let 5x-6-x²=t

⇒ (5-2x) dx=dt.

When x=2, t=0 and when x=3, y=0.

Hence

Let,

=π

Hence,

Let

Using , we get

Let

⇒ .

Also, when ,

and when ,

Let

Let x³=t

⇒ 3x²=dt.

Also, when x=0, t=0 and when , .

Hence,

Let

Let

⇒ .

Also, when x=1, t=1 and when x=2,

Hence

Let

Let sinx=t

⇒ cos x dx=dt.

Also, when , and when , t=1.

(Using )

Let, sin x + cos x = t

⇒ (cos x – sin x) dx = dt

At x = 0, t = 1

At x = π/2, t = 1

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

y = 0

Use King theorem of definite integral

Use King theorem of definite integral

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

Let, cos x = t

⇒ -sin x dx = dt

At x = 0, t = 1

At x = π, t = -1

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

Let, cos x = t

⇒ -sin x dx = dt

At x = 0, t = 1

At x = π, t = -1

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

We break it in two parts

Let, tan x = t

⇒ sec²x dx = dt

At x = 0, t = 0

At x = π, t = 0

We know that when upper and lower limit is same in definite

integral then value of integration is 0.

So, y = 0

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

[Use cot x tan x = 1]

Let, x = tan t

⇒ dx = sec²t dt

At x = 0, t = 0

At x = ∞, t = π/2

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

Let, x = a sin t

⇒ dx = a cos t dt

At x = 0, t = 0

At x = a, t = π/2

Again, sin t + cos t = z

⇒ (cos t – sin t) dt = dz

At t = 0, z = 1

At t = π/2, z = 1

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

y = 0

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

y = 0

Let, sin x + cos x = t

⇒ cos x – sin x dx = dt

At x = 0, t = 1

At x = π/2, t = 1

We know that when upper and lower limit in definite integral is

equal then value of integration is zero.

So, y = 0

Let, …(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

Let, 2x = t

⇒ 2 dx = dt

At x = 0, t = 0

At x = π/2, t = π

Similarly,

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

…(3)

Use King theorem of definite integral

…(4)

Adding eq.(3) and eq.(4)

Let, 2x = t

⇒ 2 dx = dt

At x = 0, t = 0

At x = π/2, t = π

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

…(3)

Use King theorem of definite integral

…(4)

Adding eq.(3) and eq.(4)

Let, 2x = t

⇒ 2 dx = dt

At x = 0, t = 0

At x = π/2, t = π

Let, …(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

Let, 2x = t

⇒ 2 dx = dt

At x = 0, t = 0

At x = π/2, t = π

Similarly,

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

Let, cos x = t

⇒ -sin x dx = dt

At x = π/4, t =

At x = 3π/4, t =

…(1)

Use King theorem of definite integral

Adding eq.(1) and eq.(2)

Use King theorem of definite integral

Adding eq.(1) and eq.(2)

Use integration by parts

Let, …(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

Let, 2x = t

⇒ 2 dx = dt

At x = 0, t = 0

At x = π/2, t = π

Let, x = sin t

⇒ dx = cos t dt

At x = 0, t = 0

At x = 1, t = π/2

Use integration by parts

Let, …(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

Let, 2t = z

⇒ 2 dt = dz

At t = 0, z = 0

At t = π/2, z = π

Use integration by parts

Let, x = sin t

⇒ dx = cos t dt

At x = 0, t = 0

At x = 1, t = π/2

Use integration by parts

Let, …(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

Let, 2t = z

⇒ 2 dt = dz

At t = 0, z = 0

At t = π/2, z = π

Let x = tan t

⇒ dx = sec²t dt

At x = 0, t = 0

At x = 1, t = π/4

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

y = 0

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

y = 0

…(1)

Use King theorem of definite integral

…(2)

Adding eq.(1) and eq.(2)

y = 0

We know that

|x| = -x in [-1, 0)

|x| = x in [0, 1]

y= -(1-e)+(e-1)

y = 2(e – 1)

We know that

|x+1| = -(x+1) in [-2, -1)

|x+1| = (x+1) in [-1, 2]

=5

We know that

|x – 5| = -(x – 5) in [0, 5)

|x – 5| = (x – 5) in [5, 8]

=17

We know that

|cos x| = cos x in [0, π/2)

|cos x| = -cos x in [π/2, 3π/2)

|cos x| = cos x in [3π/2, 2π]

y=(1-0)—1-1+(0+1)

=4

We know that

|sin x| = -sin x in [-π/4, 0)

|sin x| = sin x in [0, π/4]