The integral ∫ π 6 π 3 s e c 2 3 x · c o s e c 4 3 x d x is equal to

Step-by-step solution

Let I = ∫ π 6 π 3 sec 2 3 x · cosec 4 3 x d x ⇒ I = ∫ π 6 π 3 1 cos 2 3 x · sin 4 3 x d x ⇒ I = ∫ π 6 π 3 cos 4 3 x cos 2 3 + 4 3 x · sin 4 3 x d x ⇒ I = ∫ π 6 π 3 sec 2 x tan 4 3 x d x Let tan x = t ,  sec 2 x d x = d t and at x = π 6 ,   t = tan π 6 = 1 3 and at x = π 3 ,   t = tan π 3 = 3 ⇒ I = ∫ 1 3 3 d t t 4 3 Using ∫ x n d x = x n + 1 n + 1 ⇒ I = t - 4 3 +