Question

\[\int\limits_0^1 \frac{d}{dx}\left\{ \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \right\} dx\] is equal to

विकल्प

•  0

•  π

• π/2

• π/4

Answer 1

π/2

\[\text{We have}, \]

\[I = \int_0^1 \frac{d}{dx}\left\{ \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \right\}dx\]

\[\text{We know since} \int f'(x) = f(x)\]

\[f(x) = si n^{- 1} \left( \frac{2x}{1 + x^2} \right) and f'(x) = \frac{d}{dx}\left\{ si n^{- 1} \left( \frac{2x}{1 + x^2} \right) \right\} \]

\[\text{Therefore}, I = \left[ \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \right]_0^1 \]

\[ = \sin^{- 1} \left( 1 \right) - \sin^{- 1} \left( 0 \right)\]

\[ = \frac{\pi}{2}\]