Question

${\int }_0^{\pi }\frac{xtanx}{secx+tanx}d{x}_{=}$

• A. $\frac{\pi -2}{2}$
• B. $\frac{\pi (\pi -2)}{2}$
• C. $\frac{\pi +2}{2}$
• D. $\frac{\pi (\pi +2)}{2}$

Answer 1

The correct option is B $\frac{\pi (\pi -2)}{2}$

Let $I={\int }_0^{\pi }\frac{xtanx}{tanx+secx}dx$ ...(1)
Using property ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$
We get
$I={\int }_0^{\pi }\frac{(\pi -x)tan(\pi -x)}{tan(\pi -x)+sec(\pi -x)}dx={\int }_0^{\pi }\frac{(\pi -x)tanx}{tanx+secx}dx$ ...(2)
Adding (1) and (2)
$2I={\int }_0^{\pi }\frac{\pi tanx}{tanx+secx}dx$
Substitute $tan\left(\frac{x}{2}\right)=t\Rightarrow \frac{1}{2}{sec}^2\left(\frac{x}{2}\right)dx=dttanx=\frac{2t}{1-t^2},secx=\frac{1+t^2}{1-t^2},dx=\frac{2}{1+t^2}$
We get
$2I=4\pi {\int }_0^{\infty }\frac{tdt}{{(t+1)}^2(t^2+1)}=2\pi {\left[\frac{(t+1){tan}^{-1}t+1}{t+1}\right]}_0^{\infty }=\pi {[x-\frac{2sin\left(\frac{x}{2}\right)}{cos\left(\frac{x}{2}\right)+sin\left(\frac{x}{2}\right)}]}_0^{\pi }=\frac{\pi (\pi -2)}{2}$