Evaluate ∫ x e x ( 1 + x ) 2 d x

To find: Value of ∫ e x x 1 + x 2 d x Formula used: ∫ e x f ( x ) + f ' ( x ) d x = e x f ( x ) + c We have, I = ∫ e x x 1 + x 2 d x ....(i) I = ∫ e x x + 1 - 1 1 + x 2 d x I = ∫ e x 1 1 + x - 1 1 + x 2 d x Here, f ( x ) = 1 1 + x f ' x = - 1 1 + x 2 Therefore, I = e x 1 1 + x + c

Evaluate ∫ x e x ( 1 + x ) 2 d x

To find: Value of ∫ e x x 1 + x 2 d x Formula used: ∫ e x f ( x ) + f ' ( x ) d x = e x f ( x ) + c We have, I = ∫ e x x 1 + x 2 d x ....(i) I = ∫ e x x + 1 - 1 1 + x 2 d x I = ∫ e x 1 1 + x - 1 1 + x 2 d x Here, f ( x ) = 1 1 + x f ' x = - 1 1 + x 2 Therefore, I = e x 1 1 + x + c

E v a l u a t e : ∫ 0 1 tan - 1 x 1 + x 2 d x

s o l v e , ∫ 0 1 tan - 1 x 1 + x 2 d x tan - 1 x = t 1 1 + x 2 d x = d t tan - 1 x = t tan - 1 ( 1 ) = t tan - 1 ( tan π 4 ) = t π 4 = t tan - 1 x = t tan - 1 ( tan 0 ) = t 0 = t = ∫ 0 π 4 t d t = t 2 2 0 π 4 = 1 2 π 4 2 - 0 = 1 2 π 2 16 = π 2 32

Evaluate: ∫ sin ⁡ tan − 1 ⁡ x 1 + x 2 d x , x ∈ R .

Let I = ∫ sin ( tan − 1 x ) 1 + x 2 d x Putting tan − 1 x = t ⇒ 1 1 + x 2 = d t d x ⇒ d x 1 + x 2 = d t So, I = ∫ sin t d t ⇒ I = - cos t + c ⇒ I = - cos ( tan - 1 x ) + c Thus, we get ∫ sin ( tan − 1 x ) 1 + x 2 d x = - cos ( tan - 1 x ) + c , where c is an integration constant.

Evaluate : ∫ ( 1 - x ) x d x

Given, ∫ ( 1 - x ) x d x = ∫ x 1 2 - x 3 2 d x = ∫ x 1 2 d x - ∫ x 3 2 d x = x 3 2 3 2 - x 5 2 5 2 + c ∴ ∫ ( 1 - x ) x d x = 2 3 x 3 2 - 2 5 x 5 2 + c .

Evaluate ∫ 0 π 2 sin 4 x sin 4 x + cos 4 x d x .

Let I = ∫ 0 π 2 sin 4 x sin 4 x + cos 4 x d x .........(i) We know that, ∫ 0 a f ( x ) d x = ∫ 0 a f ( a - x ) d x I = ∫ 0 π 2 sin 4 π 2 - x sin 4 π 2 - x + cos 4 π 2 - x d x I = ∫ 0 π 2 cos 4 x cos 4 x + sin 4 x d x ......(ii) Add (i) and (ii), 2 I = ∫ 0 π 2 sin 4 x + cos 4 x sin 4 x + cos 4 x d x ⇒ 2 I = ∫ 0 π 2 d x ⇒ 2 I = [ x ] 0 π / 2 ⇒ I = π 4 Thus, ∫ 0 π 2 sin 4 x sin 4 x + cos 4 x d x = π 4 .

Evaluate ∫ x 2 + 1 e x ( x + 1 ) 2 d x .

Let I = ∫ x 2 + 1 e x ( x + 1 ) 2 d x = ∫ x 2 + 1 + 2 x - 2 x ( x + 1 ) 2 e x d x = ∫ ( x + 1 ) 2 e x ( x + 1 ) 2 d x - ∫ 2 x ( x + 1 ) 2 e x d x = ∫ e x d x - 2 ∫ ( x + 1 - 1 ) e x ( x + 1 ) 2 d x = e x - 2 ∫ ( x + 1 ) e x ( x + 1 ) 2 d x + 2 ∫ e x ( x + 1 ) 2 d x = e x - 2 ∫ e x ( x + 1 ) d x + 2 ∫ e x ( x + 1 ) 2 d x .........(i) Let I 1 = ∫ e x ( x + 1 ) d x , I 2 = ∫ e x ( x + 1 ) 2 d x Now, I 1 = ∫ e x ( x + 1 ) = 1 x + 1 ∫ e x - ∫ d d x 1 x + 1 ∫ e x d x d x = e x x + 1 - ∫ - 1 ( x + 1 ) 2 e x d x = e x x + 1 + ∫ e x ( x + 1 ) 2 d x ⇒ I 2 - I 1 = - e x x + 1 Putting this in equation (i), we get I = e x + 2 I 2 - I 1 Thus, the solution is I = e x - 2 e x x + 1 + C

Evaluate ∫ x 2 + 1 e x ( x + 1 ) 2 d x .

Let I = ∫ x 2 + 1 e x ( x + 1 ) 2 d x = ∫ x 2 + 1 + 2 x - 2 x ( x + 1 ) 2 e x d x = ∫ ( x + 1 ) 2 e x ( x + 1 ) 2 d x - ∫ 2 x ( x + 1 ) 2 e x d x = ∫ e x d x - 2 ∫ ( x + 1 - 1 ) e x ( x + 1 ) 2 d x = e x - 2 ∫ ( x + 1 ) e x ( x + 1 ) 2 d x + 2 ∫ e x ( x + 1 ) 2 d x = e x - 2 ∫ e x ( x + 1 ) d x + 2 ∫ e x ( x + 1 ) 2 d x .........(i) Let I 1 = ∫ e x ( x + 1 ) d x , I 2 = ∫ e x ( x + 1 ) 2 d x Now, I 1 = ∫ e x ( x + 1 ) = 1 x + 1 ∫ e x - ∫ d d x 1 x + 1 ∫ e x d x d x = e x x + 1 - ∫ - 1 ( x + 1 ) 2 e x d x = e x x + 1 + ∫ e x ( x + 1 ) 2 d x ⇒ I 2 - I 1 = - e x x + 1 Putting this in equation (i), we get I = e x + 2 I 2 - I 1 Thus, the solution is I = e x - 2 e x x + 1 + C

Evaluate ∫ 0 π 2 sin 4 x sin 4 x + cos 4 x d x .

Let I = ∫ 0 π 2 sin 4 x sin 4 x + cos 4 x d x .........(i) We know that, ∫ 0 a f ( x ) d x = ∫ 0 a f ( a - x ) d x I = ∫ 0 π 2 sin 4 π 2 - x sin 4 π 2 - x + cos 4 π 2 - x d x I = ∫ 0 π 2 cos 4 x cos 4 x + sin 4 x d x ......(ii) Add (i) and (ii), 2 I = ∫ 0 π 2 sin 4 x + cos 4 x sin 4 x + cos 4 x d x ⇒ 2 I = ∫ 0 π 2 d x ⇒ 2 I = [ x ] 0 π / 2 ⇒ I = π 4 Thus, ∫ 0 π 2 sin 4 x sin 4 x + cos 4 x d x = π 4 .

Evaluate : ∫ ( 1 - x ) x d x

Given, ∫ ( 1 - x ) x d x = ∫ x 1 2 - x 3 2 d x = ∫ x 1 2 d x - ∫ x 3 2 d x = x 3 2 3 2 - x 5 2 5 2 + c ∴ ∫ ( 1 - x ) x d x = 2 3 x 3 2 - 2 5 x 5 2 + c .

Evaluate: ∫ sin ⁡ tan − 1 ⁡ x 1 + x 2 d x , x ∈ R .

Let I = ∫ sin ( tan − 1 x ) 1 + x 2 d x Putting tan − 1 x = t ⇒ 1 1 + x 2 = d t d x ⇒ d x 1 + x 2 = d t So, I = ∫ sin t d t ⇒ I = - cos t + c ⇒ I = - cos ( tan - 1 x ) + c Thus, we get ∫ sin ( tan − 1 x ) 1 + x 2 d x = - cos ( tan - 1 x ) + c , where c is an integration constant.

E v a l u a t e : ∫ 0 1 tan - 1 x 1 + x 2 d x

s o l v e , ∫ 0 1 tan - 1 x 1 + x 2 d x tan - 1 x = t 1 1 + x 2 d x = d t tan - 1 x = t tan - 1 ( 1 ) = t tan - 1 ( tan π 4 ) = t π 4 = t tan - 1 x = t tan - 1 ( tan 0 ) = t 0 = t = ∫ 0 π 4 t d t = t 2 2 0 π 4 = 1 2 π 4 2 - 0 = 1 2 π 2 16 = π 2 32

Find the following integrals: ∫ 1 - sin x cos 2 x d x .

Given that, the integral is ∫ 1 - sin x cos 2 x d x = ∫ 1 cos 2 x - ∫ sin x cos 2 x d x = ∫ sec 2 x · d x - ∫ sin x cos 2 x d x Let cos x = t , - sin x d x = d t = tan x + ∫ d t t 2 = tan x - 1 t + c = tan x - 1 cos x + c

Find the following integrals: ∫ x 2 / 3 + 1 d x .

Given that, the integral is ∫ x 2 3 + 1 d x We know that, ∫ x n d x = x n + 1 n + 1 + c , n ≠ - 1 ⇒ ∫ x 2 3 + 1 d x = x 2 3 + 1 2 3 + 1 + x + c = 3 x 5 3 5 + x + c