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WBJEE · Maths · Indefinite Integration

\(\int \frac{\sin ^{-1} x}{\sqrt{1-x^2}} d x\) is equal to
where \(c\) is an arbitrary constant

  1. A \(\log \left(\sin ^{-1} \mathrm{x}\right)+\mathrm{c}\)
  2. B \(\frac{1}{2}\left(\sin ^{-1} x\right)^2+c\)
  3. C \(\log \left(\sqrt{1-x^2}\right)+c\)
  4. D \(\sin \left(\cos ^{-1} \mathrm{x}\right)+\mathrm{c}\)
Verified Solution

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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