Mathematics

Evaluate by using properies of definite integrals $$\int_{0}^{\pi }\frac{xdx}{a^{2}cos^{2}x+b^{2}sin^{2}x}$$ 


SOLUTION
$$I=\int^\pi_0\cfrac{x}{a^2\cos^2 x+b^2\sin^2 x}dx$$
$$\implies $$ $$I=\int^\pi_0\cfrac{\pi-x}{a^2\cos^2 x+b^2\sin^2 x}dx$$
Adding above two equations $$I+I=\int^\pi_0\cfrac{x}{a^2\cos^2 x+b^2\sin^2 x}dx$$$$+\int^\pi_0\cfrac{\pi-x}{a^2\cos^2 x+b^2\sin^2 x}dx$$
$$2I=\int^\pi_0\cfrac{\pi}{a^2\cos^2 x+b^2\sin^2 x}dx$$
$$2I=\int^\pi_0\cfrac{\pi}{a^2\cos^2 x+b^2\sin^2 x}dx$$
$$I=\cfrac{\pi}{2}\int^\pi_2\cfrac{sec^{2} x}{a^2+b^2\tan^2x}$$
Now, let $$tan$$ $$x$$ $$=$$ $$t$$
Therefore,
$$sec^{2} x dx = dt$$
$$I=\dfrac{2\pi}{2}\int^\infty_{0}\dfrac{dt}{a^{2} + b^{2}t^{2}}$$
$$I=\dfrac{\pi}{b^{2}}\int^\infty_{0}\dfrac{dt}{(\dfrac{a}{b})^{2} + t^{2}}$$
$$I=\dfrac{\pi}{b^{2}}\times \dfrac{b}{a} [tan^{-1}\dfrac{at}{b}]^\infty_{0}$$
$$I=\dfrac{\pi}{ab}\times \dfrac{\pi}{2}$$
$$I=\dfrac{\pi^{2}}{2ab}$$
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