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$\int \frac{\sin ^{-1} x}{\sqrt{1-x^2}} d x$ is equal to
where $c$ is an arbitrary constant
Options:
where $c$ is an arbitrary constant
Solution:
2919 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{2}\left(\sin ^{-1} x\right)^2+c$
$$
\text { Hints : } \begin{array}{rlr}
& I=\int t d t & \sin ^{-1} x=t \\
& =\frac{1}{2} t^2+c & \frac{1}{\sqrt{1-x^2}} d x=d t \\
& =\frac{1}{2}\left(\sin ^{-1} x\right)^2+c &
\end{array}
$$
\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}
$$
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