An integration: $$\int_{0}^{\pi} \frac{\sin^2 x}{a^2 - 2ab\cos x +b^2} \,\mathrm dx$$
I am stuck with this definite integral. Will putting $\,\cos x = z\,$ help out here?
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@tired: your examples don't help. The cosine term over $[\pi, 2 \pi]$ is not equal to itself over $[0,\pi]$. – Ron Gordon Aug 15 '15 at 11:23
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@RonGordon thanks, i somehow saw a $2 \pi$ in the upper integration limit :/ – tired Aug 15 '15 at 11:26
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u may use a weierstrass subsitution and finish it off by residue theorem! :) – tired Aug 15 '15 at 12:37
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The antiderivative calculated by Wolfram is
$$\frac{x(a^2+b^2)-2(a^2-b^2)\tan^{-1}\bigg(\frac{(a+b)\tan\left(\frac{x}{2}\right)}{a-b}\bigg)+2ab\sin(x)}{4a^2b^2}$$
For $a=b$, the intgegral becomes
$$\int_0^{\pi} \frac{\sin^2(x)}{2a^2(1-\cos(x)}\mathrm dx=\frac{\pi}{2a^2}$$
For $a>b$ the $\tan^{-1}$-term tends to $\frac{\pi}{2}$ because the argument tends to infinite for $x=\pi$. For $x=0$, the antiderivate is $0$.
The final result is $\dfrac{\pi}{2a^2}$
If $a<b$, either change the roles of $a$ and $b$ or use $-\dfrac{\pi}{2}$ for the problematical term in the antiderivate. The result is now $\dfrac{\pi}{2b^2}$
So, the integral is $\dfrac{\pi}{2\max(a,b)^2}$.
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The result $\frac{\pi}{2max(a,b)^2}$ is also symmetric in $a$ and $b$. Where is the problem ? – Peter Aug 15 '15 at 12:05
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The result $\frac{\pi}{2a^2}$ is only valid for $a\ge b$. For $a<b$, the result is $\frac{\pi}{2b^2}$. – Peter Aug 15 '15 at 12:07