Question

∫(dx/(cos³x√(2sin2x))) = (tanx)^A + C(tanx)^B + k, where k is a constant of integration, then A + B + C equals

(A) 16/5

(B) 27/10

(C) 7/10

(D) 21/15

Answer 1

Answer is (A) 16/5

We have

Put tan x = t. Then sec²x dx = dt.

Therefore,

Comparing this with the given equation, we get A = 1/2, B = 5/2 and C = 1/5.

Therefore,

A + B + C = 1/2 + 5/2 + 1/5 = 3 + 1/5 = 16/5