Let I x = ∫ x 2 x sec 2 + tan x ( x tan x + 1 ) 2 d x If I 0 = 0 , then I π 4 is equal to

Step-by-step solution

Given that I x = ∫ x 2 x   sec 2 x + tan x ( x   tan x + 1 ) 2 d x Apply Integration by parts. ∫ f x g x d x = f x ∫ g x d x - ∫ f ' x ∫ g x d x d x = x 2 ∫ x   sec 2 x + tan x ( x   tan x + 1 ) 2 d x - ∫ d x 2 d x ∫ x   sec 2 x + tan x ( x   tan x + 1 ) 2 d x d x Let x tan x + 1 = p ⇒ x   sec 2 x + tan x d p = d x = x 2 ∫ d p p 2 - ∫ 2 x ∫ d p p 2 d x = - x 2 ( x   tan x + 1 ) + ∫ 2 x x   tan