$\int_{0}^{1}{\frac{d}{dx}\ \left\lbrack \sin^{- 1}\left( \frac{2x}{1 + x^{2}} \right) \right\rbrack\ dx}$ is equal to

Question

Dr. Abubacker · Accepted Answer

(c)
$\int_{0}^{1}{\frac{d}{dx}\left\lbrack \sin^{- 1}\left( \frac{2x^{'}}{1 + x^{2}} \right) \right\rbrack dx}$
$= \left\lbrack \sin^{- 1}\left( \frac{2x}{1 + x^{2}} \right) \right\rbrack_{0}^{1} = \sin^{- 1}{(1) - \sin^{- 1}{(0) = \frac{\pi}{2}}}$

Q. $\int_{0}^{1}{\frac{d}{dx}\ \left\lbrack \sin^{- 1}\left( \frac{2x}{1 + x^{2}} \right) \right\rbrack\ dx}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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