We have to solve the following integration.
$$\int_{0}^{\pi} \frac{x}{9\cos^2{x}+\sin^2{x}} dx$$
In this I replaced $x$ by $\pi - x$
But after that, I got stuck.
Find the integral $\int_{0}^{\pi} \frac{x}{9\cos^2{x}+\sin^2{x}} dx$
2
$\begingroup$
calculus
integration
definite-integrals
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1 Answers
3
We have $$I =\int_{0}^{\pi} \frac {x }{9\cos ^2 x+\sin ^2 x} dx =\int_{0}^{\pi}\frac {\pi-x}{9\cos ^2x+\sin ^2x} dx $$ Thus $$I =\frac {\pi}{2}\int \frac {1}{9\cos ^2 x +\sin ^2 x} dx $$ $$=\frac {\pi}{2}\int_{0}^{\pi}\frac {\sec^2 x}{9+\tan ^2 x} dx $$ Hope you can finish it up by substituting for $u=\tan x $.
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0$\displaystyle \int^{\frac{\pi}{2}}_{0}\frac{\sec^2 x}{9+\tan^2 x}dx + \int^{\pi}_{\frac{\pi}{2}}\frac{\sec^2 x}{9+\tan^2 x}dx$ – 2017-01-05
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0@DURGESH TIWARI I thought the OP should at least know that. That's why I didn't give the hint. – 2017-01-05