WBJEE · Maths · Indefinite Integration
\(\int \frac{\sin ^{-1} x}{\sqrt{1-x^2}} d x\) is equal to
where \(c\) is an arbitrary constant
- A \(\log \left(\sin ^{-1} \mathrm{x}\right)+\mathrm{c}\)
- B \(\frac{1}{2}\left(\sin ^{-1} x\right)^2+c\)
- C \(\log \left(\sqrt{1-x^2}\right)+c\)
- D \(\sin \left(\cos ^{-1} \mathrm{x}\right)+\mathrm{c}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{2}\left(\sin ^{-1} x\right)^2+c\)
Step-by-step Solution
Detailed explanation
\[ \text { Hints : } \begin{array}{rlr} & I=\int t d t & \sin ^{-1} x=t \\ & =\frac{1}{2} t^2+c & \frac{1}{\sqrt{1-x^2}} d x=d t \\ & =\frac{1}{2}\left(\sin ^{-1} x\right)^2+c & \end{array} \]
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