Find the integral, using the substitution. ∫ e x e x + 2 d x , u = e x + 2

Let, I = ∫ e x e x + 2 d x Put u = e x + 2 ⇒ d u = e x d x ∴ I = ∫ u d u = u 1 + 1 2 1 + 1 2 + C = 2 3 u 3 2 + C = 2 3 e x + 2 3 2 + C

Find the integral, using the substitution. ∫ e x e x + 2 d x , u = e x + 2

Let, I = ∫ e x e x + 2 d x Put u = e x + 2 ⇒ d u = e x d x ∴ I = ∫ u d u = u 1 + 1 2 1 + 1 2 + C = 2 3 u 3 2 + C = 2 3 e x + 2 3 2 + C

Find the integral, using the substitution. ∫ 5 x 5 x + 1 d x , u = 5 x + 1

Let, I = ∫ 5 x 5 x + 1 d x Put u = 5 x + 1 ⇒ d u = 5 d x ∴ I = ∫ u - 1 5 u d u = 1 5 ∫ d u - 1 5 ∫ d u u = 1 5 u - 1 5 ln u + C = 1 5 5 x + 1 - 1 5 ln 5 x + 1 + C

Find the integral, using the substitution. ∫ x 3 x - 1 d x , u = 3 x - 1

Let, I = ∫ x 3 x - 1 d x Put u = 3 x - 1 ⇒ d u = 3 d x ∴ I = 1 3 ∫ u u + 1 3 d u = 1 9 ∫ u 3 2 d u + 1 9 ∫ u 1 2 d u = 1 9 × u 5 2 5 2 + 1 9 × u 3 2 3 2 + C = 2 45 u 5 2 + 2 27 u 3 2 + C = 2 45 3 x - 1 5 2 + 2 27 3 x - 1 3 2 + C

Given that I = ∫ 0 1 x 2 1 - x 2 d x , then show that I = ∫ 0 1 2 π sin 2 θ d θ . Hence find the exact value of I .

We have, I = ∫ 0 1 x 2 1 - x 2 d x Put x = sin θ ⇒ d x = cos θ d θ When x = 0 then θ = 0 and when x = 1 then θ = π 2 ∴ I = ∫ 0 π 2 sin 2 θ 1 - sin 2 θ × cos θ d θ = ∫ 0 π 2 sin 2 θ cos θ × cos θ d θ = ∫ 0 π 2 sin 2 θ d θ = ∫ 0 π 2 1 - cos 2 θ 2 d θ ∵ cos 2 A = 1 - 2 sin 2 A = 1 2 ∫ 0 π 2 d θ - 1 2 ∫ 0 π 2 cos 2 θ d θ = 1 2 θ 0 π 2 - 1 2 sin 2 θ 2 0 π 2 = 1 2 π 2 - 0 - 1 4 sinπ - sin 0 = π 4

Find the exact value of integral ∫ 0 1 2 π cos 3 x d x , u = sin x

We have, I = ∫ 0 π 2 cos 3 x d x = ∫ 0 π 2 cos 2 x cos x d x = ∫ 0 π 2 1 - sin 2 x cos x d x Put u = sin x ⇒ d u = cos x d x When x = 0 then u = 0 and when x = π 2 then u = 1 ∴ I = ∫ 0 1 ( 1 - u 2 ) d u = ∫ 0 1 d u - ∫ 0 1 u 2 d u = u 0 1 - u 3 3 0 1 = 1 - 0 - 1 3 - 0 = 2 3

Find the exact value of the integral ∫ 0 4 x 3 - x d x , u = 3 - x

We have, I = ∫ 0 4 x 3 - x d x Put u = 3 - x ⇒ d u = - 1 2 x d x ⇒ d x = - 2 x d u When x = 0 then u = 3 and when x = 4 then u = 1 &there4; I = - 2 ∫ 3 1 3 - u 2 u d u = 2 ∫ 1 3 u 2 - 6 u + 9 u d u = 2 ∫ 1 3 u d u - 6 ∫ 1 3 d u + 9 ∫ 1 3 d u u = 2 u 2 2 1 3 - 6 u 1 3 + 9 ln u 1 3 = 2 9 2 - 1 2 - 6 3 - 1 + 9 ln 3 - ln 1 = 2 4 - 12 + 9 ln 3 = 18 ln 3 - 16

Find the exact value of each of these integrals. ∫ 1 2 3 x 2 - x d x , u = x

Let I = ∫ 1 2 3 x 2 - x d x Put u = x ⇒ d u = 1 2 x d x ⇒ 2 d u = d x x When x = 1 then u = 1 and when x = 2 then u = 2 ∴ I = 2 ∫ 1 2 3 u 2 - u d u = - 6 ∫ 1 2 d u u 2 - 2 u + 1 - 1 = - 6 ∫ 1 2 d u u - 1 2 - 1 = 6 ∫ 1 2 d u 1 2 − u - 1 2 = 6 1 2 log 1 + u - 1 1 - ( u - 1 ) 1 2 ∵ ∫ 1 a 2 − x 2 d x = 1 2 a log ⁡ a + x a − x + c = 3 log u 2 - u 1 2 = 3 log 2 2 - 2 - log 1 2 - 1 = 3 log 2 2 - 2 × 2 + 2 2 + 2 - 0 = 3 log 2 2 + 2 2 = 3 log 2 + 1

Find the exact value of integral ∫ 0 π 2 x cos x 2 4 d x , u = cos x 2 4

Let I = ∫ 0 π 2 x cos x 2 4 d x Put u = cos x 2 4 ⇒ d u = - x 2 sin x 2 4 d x ⇒ - 2 sin x 2 4 d u = x d x ⇒ - 2 1 - u 2 d u = x d x When x = 0 then u = 1 and when x = π then u = 1 2 ∴ I = - 4 ∫ 1 1 2 u 1 - u 2 d u Put 1 - u 2 = v ⇒ - 2 u d u = d v ⇒ - 1 2 d v = u d u When u = 1 then v = 0 and when u = 1 2 then v = 1 2 ∴ I = 2 ∫ 0 1 2 d v v = 2 2 v 0 1 2 = 4 1 2 - 0 = 2 2

Find the exact value of the integral ∫ 1 6 π 1 3 π sec 2 x tan 3 x d x , u = tan x

Let I = ∫ 1 6 π 1 3 π sec 2 x tan 3 x d x Put u = tan x ⇒ d u = sec 2 x d x When x = π 3 then u = 3 and when x = π 6 then u = 1 3 ∴ I = ∫ 1 3 3 u 3 d u = u 4 4 1 3 3 = 1 4 3 4 - 1 3 4 = 1 4 9 - 1 9 = 1 4 × 80 9 = 20 9