Question

${\int }_0^{\pi }\frac{xtanx}{secx+tanx}dx.$

Answer 1

Let $I={\int }_0^{\pi }\frac{xtanx}{secx+tanx}dx$ ...(i)
$\Rightarrow I={\int }_0^{\pi }\frac{(\pi -x)tan(\pi -x)}{sec(\pi -x)+tan(\pi -x)}dx[\because {\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x\left)dx\right]\Rightarrow I={\int }_0^{\pi }\frac{(\pi -x)tanx}{secx+tanx}dx$ ...(ii)
$[\because tan(\pi -x)=-tanxandsec(\pi -x)=-secx]$
On adding Eqs. (i) and (ii), we get
$2I={\int }_0^{\pi }\frac{\pi tanx}{secx+tanx}dx=\pi {\int }_0^{\pi }\frac{\pi tanx}{secx+tanx}dx=\pi {\int }_0^{\pi }\frac{\frac{sinx}{cosx}}{\frac{1}{cosx}+\frac{sinx}{cosx}}dx=\pi {\int }_0^{\pi }\frac{sinx}{1+sinx}dx$
$=\pi {\int }_0^{\pi }\frac{sinx(1-sinx)}{(1+sinx)(1-sinx)}dx=\pi {\int }_0^{\pi }\frac{sinx(1-sinx)}{1-si{n}^{2}x}dx$
$(\because si{n}^{2}x+co{s}^{2}x=1)$
$=\pi {\int }_0^{\pi }\frac{sinx-si{n}^{2}x}{co{s}^{2}x}dx=\pi {\int }_0^{\pi }(\frac{sinx}{co{s}^2}-\frac{si{n}^{2}x}{co{s}^{2}x})dx$
$=\pi {\int }_0^{\pi }(secxtanx-ta{n}^{2}x)dx$
$=\pi {\int }_0^{\pi }\{secxtanx-(se{c}^{2}x-1\left)\right\}dx$ $[\because 1+ta{n}^{2}x=se{c}^{2}x]$
$=\pi [secx-tanx+x{]}_0^{\pi }$
$=\pi \{sec\pi -tan\pi +\pi -(sec0-tan0+0\left)\right\}=\pi (-1+\pi -1)\therefore I=\frac{\pi }{2}(\pi -2)$