Question

Evaluate the integral: ∫x sin^-1 x/√(1 - x²) dx

Answer 1

Take sin ^-1 x = t

Here x = sin t

So we get

By integration we get

= t(- cos t) – ∫1. (- cos t) dt

We get

= – t cos t + sin t + c

It can be written as

cos t = √1 – sin² t

Here we get

= – t (√1 – sin² t) + sin t + c

By replacing t as sin^-1 x

= – sin^-1 x (√1 – x²) + x + c