Question

$\int \frac{\sin ^{-1} x}{\sqrt{1-x^2}} d x$ is equal to

(a) $\log \left(\sin ^{-1} x\right)+C$

(b) $\frac{1}{2}\left(\sin ^{-1} x\right)^2+C$

(c) $\log \left(\sqrt{1-x^2}\right)+C$

(d) $\sin \left(\cos ^{-1} x\right)+C$

Answer 1

$\begin{array}{l}\text {  Put } \sin ^{-1} x=t \Rightarrow \frac{1}{\sqrt{1-x^2}} d x=d t \\\therefore \quad I=\int t d t=\frac{t^2}{2}+C \\=\frac{\left(\sin ^{-1} x\right)^2}{2}+C \\\end{array}$

So, The correct option of this question will be (B).