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# Integrate : $\int \limits_0^{\pi} \large\frac{x}{a^2 \cos ^2x + b^2 \sin ^2 x}$$dx $\begin {array} {1 1} (a)\;\frac{\pi^2}{ab} \\ (b)\;\frac{\pi^4}{2ab} \\ (c)\;\frac{\pi^4}{ab} \\ (d)\;\frac{\pi^3}{2ab} \end {array}$ Can you answer this question? ## 1 Answer 0 votes f(\pi-x)=\int \limits_0^{\pi} \large\frac{x}{a^2 \cos ^2x + b^2 \sin ^2 x}$$dx$
$f(\pi-x)=\int \limits_0^{\pi} \pi. \large\frac{1}{a^2 \cos ^2x + b^2 \sin ^2 x}-\int \limits_0^{\pi} \large\frac{x}{a^2 \cos ^2x + b^2 \sin ^2 x}$$dx f(\pi-x)+f(x)=\pi \int \limits_0^{\pi} \large\frac{1}{a^2 \cos ^2x + b^2 \sin ^2 x}$$dx$
$f(x)=\large\frac{\pi}{2} \int \limits_0^{\pi} \large\frac{1}{a^2 \cos ^2x + b^2 \sin ^2 x}$$dx f(x)=f(2. \large\frac{\pi}{2}-x)=>$$2 \int \limits_0^a f(x).dx$
$f(x)=2. \large\frac{\pi}{2} \int \limits_0^{\pi/2} \large\frac{\sec ^2 x}{a^2+b^2 \tan ^2 x}$$dx f(x)=\pi \int \limits_0^{\pi/2} \large\frac{\sec ^2 x}{a^2+b^2 \tan ^2 x}$$dx$
$\tan x =t\; \sec^2 x dx=dt$
$x=0,t=0,x = \large\frac{z}{2},$$t=\theta \qquad =\pi \int \limits_0^{\infty}$$ \large \frac{dt}{(a/b)^2+t^2}$
$\qquad= \large\frac{\pi}{b^2 (a/b)} $$\bigg[ \tan^{-1} (\large\frac{t}{a/b})\bigg]_0^{\infty} \qquad= \large\frac{\pi}{ab}$$ [\tan^{-1} (\theta)- \tan^{-1} (\theta)]$
$\qquad= \large\frac{\pi^2}{2ab}$
Hence b is the correct answer.