Question

The value of $\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\frac{\sin x}{1+{\cos }^{2}x}dx$ is ________________.

Answer 1

$$\begin{gathered} \mathrm{Let}I=\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\frac{\sin x}{1+{\cos }^{2}x}dx \\ \mathrm{Let}\cos x=t \\ \Rightarrow d\left(\cos x\right)=dt \\ \Rightarrow -\sin xdx=dt \\ \mathrm{Also},\mathrm{if}x=0,t=1 \\ \mathrm{if}x=\frac{\mathrm{\pi }}{2},t=0 \\ \mathrm{Thus}, \\ I=\underset{1}{\overset{0}{\int }}\frac{-1}{1+t^2}dt \\ =-{\left[{\tan }^{-1}t\right]}_1^0 \\ =-\left[{\tan }^{-1}0-{\tan }^{-1}1\right] \\ =-\left[0-\frac{\mathrm{\pi }}{4}\right] \\ =\frac{\mathrm{\pi }}{4} \end{gathered}$$

​Hence, the value of $\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\frac{\sin x}{1+{\cos }^{2}x}dx$ is $\underline{\frac{\mathrm{\pi }}{4}}$.