Question

Integrate the function.
$\int (si{n}^{-1}x{)}^{2}dx.$

Answer 1

Let $I=\int (si{n}^{-1}x{)}^{2}dx$
Put $si{n}^{-1}x=\theta \Rightarrow x=sin\theta \Rightarrow dx=cos\theta d\theta$
$\therefore I=\int (si{n}^{-1}x{)}^{2}dx=\int {\theta }^{2}cos\theta d\theta$
On taking ${\theta }^2$ as first function and cos $\theta$ as second function and integrating by parts, we get
$={\theta }^2\int cos\theta d\theta -\int \left[\frac{d}{d\theta }\right(\theta {)}^2\int cos\theta d\theta ]d\theta ={\theta }^{2}sin\theta -\int 2\theta sin\theta d\theta$
Again integrating by parts, we get
$I={\theta }^{2}sin\theta -2\left[\theta \right(-cos\theta )-\int 1(-cos\theta \left)d\theta \right]+C={\theta }^{2}sin\theta +2\theta cos\theta -2\int cos\theta d\theta +C={\theta }^{2}sin\theta +2\theta cos\theta -2sin\theta +C=(si{n}^{-1}x{)}^{2}x+2si{n}^{-1}x\sqrt{1-si{n}^2\theta }-2x+C[Put\theta =si{n}^{-1}xandsin\theta =x]=x(si{n}^{-1}x{)}^2+2\sqrt{1-x^2}si{n}^{-1}x-2x+C$