# Definite integration evaluation of $\int_{0}^{\pi}\frac{x}{(a^2\cos^2x+b^2\sin^2x)^2}dx$

OK, so the question says evaluate the integral $$\int_{0}^{\pi}\frac{x}{(a^2\cos^2x+b^2\sin^2x)^2}dx$$ What I do is use the property that $\int_a^bf(x)dx=\int_a^bf(b+a-x)dx$ and this gives me ($I$ is the value of the integral) $$\frac{2I}{\pi}=\int_{0}^{\pi}\frac{1}{(a^2\cos^2x+b^2\sin^2x)^2}dx$$ What should I do ahead to get the value I need? Any tips? (Thanks in advance)

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Try substitution of $t = \tan(\frac{x}{2})$? –  Memming Feb 23 '13 at 17:32
Don't you think converting to half angles would complicate the question? Did you mean the tan part? I have tried $t=tan(x)$, that solves the integral (indefinite way) but on inserting the limits, I get an indeterminate answer. –  Ashish Gaurav Feb 23 '13 at 17:37

I prefer to the following method:

\begin{align*} I := \int_{0}^{\pi} \frac{x}{(a^2 \cos^2 x + b^2 \sin^2 x )^2} \, dx &= \frac{\pi}{2} \int_{0}^{\pi} \frac{dx}{(a^2 \cos^2 x + b^2 \sin^2 x )^2} \\ &= \pi \int_{0}^{\frac{\pi}{2}} \frac{dx}{(a^2 \cos^2 x + b^2 \sin^2 x )^2} \\ &= \pi \int_{0}^{\frac{\pi}{2}} \frac{1 + \tan^2 x}{(a^2 + b^2 \tan^2 x )^2} \, \sec^2 x \, dx. \end{align*}

Now we make the substitution $b \tan x \mapsto a \tan x$. Then

\begin{align*} I &= \frac{\pi}{(ab)^3} \int_{0}^{\frac{\pi}{2}} \frac{b^2 + a^2\tan^2 x}{(1 + \tan^2 x )^2} \, \sec^2 x \, dx \\ &= \frac{\pi}{(ab)^3} \int_{0}^{\frac{\pi}{2}} ( b^2 \cos^2 x + a^2\sin^2 x ) \, dx \\ &= \frac{\pi}{(ab)^3} \cdot \frac{\pi}{4} \left( a^2 + b^2 \right) = \frac{\pi^2(a^2 + b^2)}{4(ab)^3}. \end{align*}

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In the third step(part 1) where you convert the upper limit to $\frac{\pi}{2}$, there, what did you use? Did you use the symmetry of the function $f(x)=a^2cos^2x+b^2sin^2x$, that $f(\frac{\pi}{2}+t)=f(\frac{\pi}{2}-t)$? –  Ashish Gaurav Feb 23 '13 at 17:45
@AshishGaurav: Yes, absolutely right. That was because of the symmetry of the integrand that you pointed out. –  sos440 Feb 23 '13 at 17:48
Thank you so much! –  Ashish Gaurav Feb 23 '13 at 17:49
I think you also used that $$\int_0^\pi xf(\sin x)dx=\frac \pi 2 \int_0^\pi f(\sin x)dx$$ –  Pedro Tamaroff Feb 23 '13 at 17:59
@PeterTamaroff: That's also right, though I skipped the explanation because OP was already well aware of that. –  sos440 Feb 23 '13 at 18:25