calculus - How to evaluate $int sin^{-1}(x)cos^{-1}(x) , dx$?

I need to evaluate

$$\int \sin^{-1}(x)\cos^{-1}(x) \, dx.$$ Can anyone please give me an idea or a hint ? Thanks.

Answer

we will use the fact that $\sin^{-1} x + \cos ^{-1} x = \pi/2$ and a change of variable $\sin^{-1} x = t, x = \sin t, dx = \cos t \, dt$ with these we get $\begin{align}\int \sin^{-1}(x)\cos^{-1}(x) \, dx &= \int t(\pi/2 - t)\cos t \, dt \\ &=\int (\pi/2 t - t^2) \, d \cos t \\ &= (\pi/2 t - t^2)\cos t - \int (\pi/2 - 2t) \cos t\, dt \\ &= (\pi/2 t - t^2)\cos t + (\pi/2 - 2t) \sin t - 2 \int \sin t\, dt\\ &= (\pi/2 t - t^2)\cos t + (\pi/2 - 2t) \sin t + 2 \cos t +C\\\end{align}$