Question

If $I={\int }_0^{\pi }\frac{x\sin x}{1+{\cos }^{2}x}dx$, then the value of $\sin \sqrt{I}$, is

• A. $\frac{1}{2}$
• B. $0$
• C. $\frac{\sqrt{2}}{2}$
• D. $1$

Answer 1

The correct option is D $1$

$I={\int }_0^n\frac{x\sin x}{1+{\cos }^{2}x}dx....\left(1\right)$

$={\int }_0^n\frac{(n-x)\sin (n-x)}{1+{\cos }^{2}x}dx$

$I=n{\int }_0^n\frac{\sin \left(x\right)}{1+{\cos }^{2}x}dx-{\int }_0^n\frac{x\sin \left(x\right)}{1+{\cos }^{2}x}dx......\left(2\right)$

$\left(1\right)$ & $\left(2\right)$

$2I=n{\int }_0^n\frac{\sin x}{1+{\cos }^{2}x}dx$

$I=\frac{n}{2}{[-{\tan }^{-1}(\cos x\left)\right]}_0^n$

$I=\frac{n}{2}[-{\tan }^{-1}(-1)+{\tan }^{-1}(1)$

$=\frac{n}{2}\left[2{\tan }^{-1}1\right]$

$=n\times \frac{n}{4}=\frac{n^2}{4}$

$\sin \sqrt{I}=\sin \frac{n}{2}=1$

comment : Correct answer $D)1$