What is the integration of sin inverse x whole square ?

Solution :

We have, I = \((sin^{-1}x)^2\) dx

Let \(sin^{-1}x\) = t, Then, x = sin t \(\implies\) dx = cos t dt

\(\therefore\) I = \(\int\) \((sin^{-1}x)^2\) dx

I = \(\int\) \(t^2\) cos t dt

Applying integration by parts and,

Taking \(t^2\) as first function and cos t as second function,

I = \(t^2\) (sin t) – \(\int\) (sin t) – \(\int\) 2t sin t dt = \(t^2\) sin t – 2 \(\int\) t sin t dt

Again applying integration by parts and,

Taking t as first function and sin t as second function,

I = \(t^2\) sin t – 2{ t ( -cos t) – \(\int\) 1 \(\times\) (-cos t) dt }

I = \(t^2\) sin t – 2{ -t cos t + \(\int\) cos t dt }

I = \(t^2\) sin t – 2( -t cos t + sin t ) + C

I = \(t^2\) sin t – 2{ -t \(\sqrt{1 – sin^2 t}\) + sin t } + C

I = \(xsin^{-1} x\) – 2{ – \(\sqrt{1 – x^2} sin^{-1}x\) + x } + C