Question

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

Answer 1

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

$⟹$ $I={\int }_0^{\pi }\frac{\pi -x}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}dx$

Adding above two equations $I+I={\int }_0^{\pi }\frac{x}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}dx$$+{\int }_0^{\pi }\frac{\pi -x}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}dx$

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

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

$I=\frac{\pi }{2}{\int }_0^{\pi }\frac{se{c}^{2}x}{a^2+b^2{\tan }^{2}x}$

Now, let $tan$ $x$ $=$ $t$

Therefore,

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

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

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

$I=\frac{\pi }{b^2}\times \frac{b}{a}[ta{n}^{-1}\frac{at}{b}{]}_0^{\infty }$

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

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