Question

Evaluate the integral: ∫x tan² x dx

Answer 1

It is given that

∫x tan² x dx

We can write it as

tan² x = sec² x – 1

So we get

∫xtan² x dx = ∫x( sec² x – 1) dx

By further simplification

= ∫ x sec² x dx – ∫ x dx

Taking first function as x and second function as sec² x

∫ x sec² x dx – ∫ x dx

We get

By integration w.r.t x

= {x tan x – ∫tan x dx} – x²/ 2

It can be written as

= x tan x – log |sec x| – x²/ 2 + c

So we get

= x tan x – log |1/cos x| – x²/ 2 + c

Here

= x tan x + log |cos x| – x²/ 2 + c