Question

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

Answer 1

${\int }_0^{\pi }\frac{xdx}{a^{2}co{s}^{2}x+b^{2}si{n}^{2}x}$

$I={\int }_0^{\pi }\frac{xdx}{a^{2}co{s}^{2}x+b^{2}si{n}^{2}x}$

$I={\int }_0^{\pi }\frac{(\pi -x)dx}{a^{2}co{s}^{2}x+b^{2}si{n}^{2}x}$

$2I=\pi {\int }_0^{\pi }\frac{1}{a^{2}co{s}^{2}x+b^{2}si{n}^{2}x}dx$

$2I=2\pi {\int }_0^{\pi /2}\frac{se{c}^{2}x}{a^2+b^{2}ta{n}^{2}x}dx$

b tan x = t

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

$I=\frac{\pi }{6}\int \frac{dt}{a^2+t^2}$

$I=\frac{\pi }{6}{\int }_0^{\infty }\frac{dt}{a^2+t^2}$

$I=\frac{\pi }{ab}(ta{n}^{-1}\frac{x}{a}{)}_0^{\infty }$

$I=\frac{{\pi }^2}{2ab}$