Question

${\int }_0^{\pi }\frac{xdx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}=\frac{{\pi }^2}{2ab}$

Answer 1

$I={\int }_o^{\pi }\frac{xdx}{a^{2}co{s}^{2}x+b^{2}si{n}^{2}x}$...(1)

$I={\int }_o^{\pi }\frac{(\pi -x)dx}{a^{2}co{s}^{2}x+b^{2}si{n}^{2}x}$..(2)

Adding (1) & (2)

$2I={\int }_o^{\pi }\frac{x+\pi -x}{a^{2}co{s}^{2}x+b^{2}si{n}^{2}x}$

$2I={\int }_o^{\pi }\frac{\pi dx}{a^{2}co{s}^{2}x+b^{2}si{n}^{2}x}$

$={\int }_o^{\pi }\frac{\pi se{c}^{2}xdx}{a^2+b^{2}ta{n}^{2}x}=\frac{1}{b^2}{\int }_o^{\pi }\frac{\pi se{c}^{2}xdx}{\left(\frac{a}{b}\right)+ta{n}^x}$

Let $tanx=tat$

$se{c}^{2}xdx=dt$

$2I=\frac{\pi }{b^2}{\int }_0^{\pi }\frac{dt}{(\frac{a}{b}{)}^2+t^2}$

$=\frac{\pi }{b^2}\times \frac{a}{b}\times \left[ta{n}^{-1}t\right]$

$2I=\frac{\pi }{ab}\left[ta{n}^{-1}\right(tanx){]}_0^{\pi }$

$2I=\frac{\pi }{ab}[x{]}_0^{\pi }$

$I=\frac{\pi }{ab}\left[x\right]=\frac{{\pi }^2}{2ab}$