Question

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

• A. 0
• B. $\pi$
• C. $\pi ⥂ / ⥂ 2$
• D. $\pi ⥂ / ⥂ 4$

Answer 1

Answer: (C)

$\pi ⥂ / ⥂ 2$

Answer 2

$\mathbf{I =}\left\lbrack \mathbf{\sin}^{\mathbf{- 1}}\left( \frac{\mathbf{2x}}{\mathbf{1 +}\mathbf{x}^{\mathbf{2}}} \right) \right\rbrack_{\mathbf{0}}^{\mathbf{1}}\mathbf{=}\mathbf{\sin}^{\mathbf{- 1}}\mathbf{(}\mathbf{1) -}\mathbf{\sin}^{\mathbf{- 1}}\mathbf{(}\mathbf{0)}\mathbf{=}\frac{\mathbf{\pi}}{\mathbf{2}}$.