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प्रश्न
पर्याय
0
π
π/2
π/4
उत्तर
π/2
\[\text{We have}, \]
\[I = \int_0^1 \frac{d}{dx}\left\{ \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \right\}dx\]
\[\text{We know since} \int f'(x) = f(x)\]
\[f(x) = si n^{- 1} \left( \frac{2x}{1 + x^2} \right) and f'(x) = \frac{d}{dx}\left\{ si n^{- 1} \left( \frac{2x}{1 + x^2} \right) \right\} \]
\[\text{Therefore}, I = \left[ \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \right]_0^1 \]
\[ = \sin^{- 1} \left( 1 \right) - \sin^{- 1} \left( 0 \right)\]
\[ = \frac{\pi}{2}\]
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