Evaluate the following definite integrals:
Question 21.
Solution:
We have,
I =
Let sin x = A (sin x + cos x) + B
=> sin x = A (sin x + cos x) + B (cos x – sin x)
=> sin x = sin x (A – B) + cos x (A + B)
On comparing both sides, we get
A – B = 1 and A + B = 0
On solving, we get A = 1/2 and B = –1/2.
Therefore, the expression becomes,
I =
I =
I =
I =
Therefore, the value of
is .
Question 22.
Solution:
We have,
I =
On putting cos x =
and sin x = , we get, I =
I =
I =
I =
Let tan x/2 = t. So, we have
=> 1/2 sec2 x/2 dx = dt
=> sec2 x/2 dx = 2 dt
Now, the lower limit is, x = 0
=> t = tan x/2
=> t = tan 0/2
=> t = tan 0
=> t = 0
Also, the upper limit is, x = π
=> t = tan x/2
=> t = tan π/2
=> t = ∞
So, the equation becomes,
I =
I =
I =
I =
I =
I =
I =
Therefore, the value of
is .
Question 23.
Solution:
We have,
I =
I =
I =
I =
I =
I =
I =
I =
I =
Therefore, the value of
is .
Question 24.
Solution:
We have,
I =
Let sin–1 x = t. So, we have
=>
= dt Now, the lower limit is, x = 0
=> t = sin–1 x
=> t = sin–1 0
=> t = 0
Also, the upper limit is, x = 1/2
=> t = sin–1 x
=> t = sin–1 1/2
=> t = π/6
So, the equation becomes,
I =
I =
I =
I =
I =
I =
I =
Therefore, the value of
is .
Question 25.
Solution:
We have,
I =
I =
I =
I =
I =
Let sinx – cosx = t. So, we have
=> (cos x + sin x) dx = dt
Now, the lower limit is, x = 0
=> t = sinx – cosx
=> t = sin 0 – cos 0
=> t = 0 – 1
=> t = –1
Also, the upper limit is, x = π/4
=> t = sinx – cosx
=> t = sin π/4 – cos π/4
=> t = sin π/4 – sin π/4
=> t = 0
So, the equation becomes,
I =
I =
I =
I =
I =
I =
Therefore, the value of
is .
Question 26.
Solution:
We have,
I =
I =
I =
I =
Let tan x = t. So, we have
=> sec2 x dx = dt
Now, the lower limit is, x = 0
=> t = tan x
=> t = tan 0
=> t = 0
Also, the upper limit is, x = π/4
=> t = tan x
=> t = tan π/4
=> t = 1
So, the equation becomes,
I =
I =
I =
I =
Therefore, the value of
is .
Question 27.
Solution:
We have,
I =
On putting cos x =
, we get I =
I =
I =
Let tan x/2 = t. So, we have
=> 1/2 sec2 x/2 dx = dt
=> sec2 x/2 dx = 2 dt
Now, the lower limit is, x = 0
=> t = tan x/2
=> t = tan 0/2
=> t = tan 0
=> t = 0
Also, the upper limit is, x = π
=> t = tan x/2
=> t = tan π/2
=> t = ∞
So, the equation becomes,
I =
I =
I =
I =
I =
I =
I =
I =
I =
Therefore, the value of
is .
Question 28.
Solution:
We have,
I =
I =
I =
I =
Let tan x = t. So, we have
=> sec2 x dx = dt
Now, the lower limit is, x = 0
=> t = tan x
=> t = tan 0
=> t = 0
Also, the upper limit is, x = π/2
=> t = tan x
=> t = tan π/2
=> t = ∞
So, the equation becomes,
I =
I =
I =
I =
I =
Therefore, the value of
is .
Question 29.
Solution:
We have,
I =
I =
I =
I =
I =
I =
I =
Therefore, the value of
is .
Question 30.
Solution:
We have,
I =
Let tan–1 x = t. So, we have
=>
= dt Now, the lower limit is, x = 0
=> t = tan–1 x
=> t = tan–1 0
=> t = 0
Also, the upper limit is, x = 1
=> t = tan–1 x
=> t = tan–1 1
=> t = π/4
So, the equation becomes,
I =
I =
I =
I =
Therefore, the value of
is .
Question 31.
Solution:
We have,
I =
I =
I =
I =
I =
I =
I =
I =
I =
I =
Therefore, the value of
is .
Question 32.
Solution:
We have,
I =
On using integration by parts, we get
I =
I =
I =
I =
I =
I =
I =
I =
I =
I =
I =
Therefore, the value of
is .
Question 33.
Solution:
We have,
I =
I =
I =
I =
I =
I =
I =
I =
I =
I =
I =
I =
I =
I =
Therefore, the value of
is .
Question 34.
Solution:
We have,
I =
Let 1 + x2 = t. So, we have
=> 2x dx = dt
Now, the lower limit is, x = 0
=> t = 1 + x2
=> t = 1 + 02
=> t = 1 + 0
=> t = 1
Also, the upper limit is, x = π
=> t = 1 + x2
=> t = 1 + 12
=> t = 1 + 1
=> t = 2
So, the equation becomes,
I =
I =
I =
I =
I =
I =
I =
I =
I = 1
Therefore, the value of
is 1.
Question 35.
Solution:
We have,
I =
Let x – 4 = t3. So, we have
=> dx = 3t2 dt
Now, the lower limit is, x = 4
=> t3 = x – 4
=> t3 = 4 – 4
=> t3 = 0
=> t = 0
Also, the upper limit is, x = 12
=> t3 = x – 4
=> t3 = 12 – 4
=> t3 = 8
=> t = 2
So, the equation becomes,
I =
I =
I =
I =
I =
I =
Therefore, the value of
is .
Question 36.
Solution:
We have,
I =
On using integration by parts, we get
I =
I =
I =
I =
I =
I = π + 0 – 0 – 0 – 2
I = π – 2
Therefore, the value of
is π – 2.
Question 37.
Solution:
We have,
I =
Let x = cos 2t. So, we have
=> dx = – 2 sin 2t dt
Now, the lower limit is, x = 0
=> cos 2t = x
=> cos 2t = 0
=> 2t = π/2
=> t = π/4
Also, the upper limit is, x = 1
=> cos 2t = x
=> cos 2t = 1
=> 2t = 0
=> t = 0
So, the equation becomes,
I =
I =
I =
I =
I =
I =
I =
I =
I =
Therefore, the value of
is .
Question 38.
Solution:
We have,
I =
I =
I =
Let x + 1/x = t. So, we have
=> (1 – 1/x2)dx = dt
Now, the lower limit is, x = 0
=> t = x + 1/x
=> t = ∞
Also, the upper limit is, x = 1
=> t = x + 1/x
=> t = 1 + 1
=> t = 2
So, the equation becomes,
I =
I =
I =
I =
I =
Therefore, the value of
is .
Question 39.
Solution:
We have,
I =
Let x5 + 1 = t. So, we have
=> 5x4 dx = dt
Now, the lower limit is, x = –1
=> t = x5 + 1
=> t = (–1)5 + 1
=> t = –1 + 1
=> t = 0
Also, the upper limit is, x = 1
=> t = x5 + 1
=> t = (1)5 + 1
=> t = 1 + 1
=> t = 2
So, the equation becomes,
I =
I =
I =
I =
I =
Therefore, the value of
is .
Question 40.
Solution:
We have,
I =
I =
Let tan x = t. So, we have
=> sec2 x dx = dt
Now, the lower limit is, x = 0
=> t = tan x
=> t = tan 0
=> t = 0
Also, the upper limit is, x = π/2
=> t = tan x
=> t = tan π/2
=> t = ∞
So, the equation becomes,
I =
I =
I =
I =
I =
Therefore, the value of
is .
Question 41.
Solution:
We have,
I =
Let sin 2t = u. So, we have
=> 2 cos 2t dt = du
=> cos 2t dt = du/2
Now, the lower limit is, x = 0
=> u = sin 2t
=> u = sin 0
=> u = 0
Also, the upper limit is, x = π/4
=> u = sin 2t
=> u = sin π/2
=> u = 1
So, the equation becomes,
I =
I =
I =
I =
Therefore, the value of
is .