Question

Solve:
$\stackrel{\pi }{\underset{0}{\int }}\frac{x\sin xdx}{1+{\cos }^{2}x}=$

• A. $\frac{{\pi }^2}{4}$
• B. $\frac{-{\pi }^2}{2}$
• C. $\frac{\pi }{3}$
• D. $-\frac{{\pi }^2}{4}$

Answer 1

The correct option is D $-\frac{{\pi }^2}{4}$

${\int }_0^{\pi }\frac{xsinxdx}{1+co{s}^{2}x}=I...\left(1\right)$

$\Rightarrow {\int }_0^{\pi }\frac{(\pi -x)sin(\pi -x)dx}{1+co{s}^2(\pi -x)}$

$\Rightarrow {\int }_0^{\pi }\frac{(\pi -x)sinx}{1+co{s}^{2}x}dx=I...\left(2\right)$

adding (1) & (2)

$2I={\int }_1^{-1}\frac{\pi sinxdx}{1+co{s}^{2}x}$

cosx = t

dt = -sinx dx

$\therefore 2I={\int }_{-1}^1\frac{\pi .dt}{1+t2}$

$2I=-\pi [ta{n}_t^{-1}{]}_{+1}^{-1}$

$I=\frac{-\pi }{2}[\pi -\frac{\pi }{4}-\frac{\pi }{4}]$

$I=\frac{-{\pi }^2}{4}$