Q86 / 160 · WBJEE 2009 · Mathematics
$\int \frac{\sin ^{-1} x}{\sqrt{1-x^2}} d x$ is equal to
where $c$ is an arbitrary constant
- A$\log \left(\sin ^{-1} \mathrm{x}\right)+\mathrm{c}$
- Correct answerB$\frac{1}{2}\left(\sin ^{-1} x\right)^2+c$
- C$\log \left(\sqrt{1-x^2}\right)+c$
- D$\sin \left(\cos ^{-1} \mathrm{x}\right)+\mathrm{c}$
Solution
$$
\text { Hints : } \begin{array}{rlr}
& I=\int t d t & \sin ^{-1} x=t \\
& =\frac{1}{2} t^2+c & \frac{1}{\sqrt{1-x^2}} d x=d t \\
& =\frac{1}{2}\left(\sin ^{-1} x\right)^2+c &
\end{array}
$$
Hence, the correct answer is option: 'B'.