Question

$\underset{0}{\overset{1}{\int }}\frac{d}{dx}\left\{{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)\right\}dx$ is equal to

(a) 0
(b) π
(c) π/2
(d) π/4

Answer 1

(c) π/2

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I={\int }_0^1\frac{d}{dx}\left\{{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)\right\}dx \\ \mathrm{We}\mathrm{know}\mathrm{since}\int f\text{'}\left(x\right)=f\left(x\right) \\ f\left(x\right)=si{n}^{-1}\left(\frac{2x}{1+x^2}\right)andf\text{'}\left(x\right)=\frac{d}{dx}\left\{si{n}^{-1}\left(\frac{2x}{1+x^2}\right)\right\} \\ \mathrm{Therefore},I={\left[{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)\right]}_0^1 \\ ={\sin }^{-1}\left(1\right)-{\sin }^{-1}\left(0\right) \\ =\frac{\mathrm{\pi }}{2} \end{gathered}$$