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I am trying to show that

$$ \int^{\pi}_{0} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx = \frac{\pi}{a(a+b)} $$ where $ a,b > 0$.

I have tried a few things, but none have worked.

For example, one approach was

$$ \int^{\pi}_{0} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx = \int^{\pi}_{0} \frac{\cos^2 x}{(a^2 - b^2)\cos^2 x + b^2} \ dx$$ and then divide by $ a^2 - b^2 $ throughout. However I am not given that $ a \not= b $, so this approach didn't work.

I also tried splitting up the integral, but then was not sure how to proceed:

$$ \int^{\pi}_{0} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx = \int^{\frac{\pi}{2}}_{0} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx + \int^{\pi}_{\frac{\pi}{2}} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx $$

Substituting $ x \to x + \frac{\pi}{2} $ followed by $ x \to \frac{\pi}{2} -x $ a in the second integral gives

$$ \int^{\frac{\pi}{2}}_{0} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx $$

Hence $$ \int^{\pi}_{0} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx = 2 \int^{\frac{\pi}{2}}_{0} \frac{\cos^2 x}{a^2 \cos^2 x + b^2 \sin^2 x} \ dx $$

Next I thought I could substitute $ u = a^2 \cos^2 x + b^2 \sin^2 x $ which changes my limits from $[0, \frac{\pi}{2}]$ to $[a^2, b^2]$, however I could not complete the substitution as I didn't see how to express $ du = (b^2 - a^2) \sin 2 x \ dx $ or $ \cos^2 x$ in terms of $u$.

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If we set $x=\arctan t$ we simply have: $$I(a,b)=\int_{0}^{\pi}\frac{\cos^2 x}{b^2\sin^2 x+a^2\cos^2 x}\,dx = 2\int_{0}^{+\infty}\frac{dt}{(1+t^2)(a^2+b^2 t^2)}$$ that is straightforward to compute through partial fraction decomposition or through the residue theorem.

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  • 1
    $\begingroup$ As a general rule: whenever one must integrate a rational function (i.e. fraction) involving even powers of $\sin$ and $\cos$, the substitution $t= \tan x$ followed by a polynomial division and then a partial fraction decomposition usually solve the problem, be it with a large number of computations. The advantage of this approach is that it is algorithmic, not requiring too much inspiration. $\endgroup$ – Alex M. Apr 11 '15 at 18:42
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Suppose we seek to evaluate $$\int_0^\pi \frac{\cos^2 x}{a^2\cos^2x + b^2\sin^2x} dx = \frac{1}{2} \int_0^{2\pi} \frac{\cos^2 x}{a^2\cos^2x + b^2\sin^2x} dx$$ with $a,b>0.$

Put $z = \exp(ix)$ so that $dz = i\exp(ix) dx$ and hence $\frac{dz}{iz} = dx$ to obtain $$\frac{1}{2}\int_{|z|=1} \frac{(z+1/z)^2}{a^2(z+1/z)^2-b^2(z-1/z)^2} \frac{dz}{iz} \\ = \frac{1}{2}\int_{|z|=1} \frac{(z^2+1)^2}{a^2(z^2+1)^2-b^2(z^2-1)^2} \frac{dz}{iz}.$$

This factors to give $$\frac{1}{2}\int_{|z|=1} \frac{(z^2+1)^2}{(a(z^2+1)+b(z^2-1))(a(z^2+1)-b(z^2-1)} \frac{dz}{iz}.$$

The roots of the first term are $$\rho_{1,2} = \pm\sqrt{\frac{b-a}{a+b}} = \pm \sqrt{1-\frac{2a}{a+b}} $$ and of the second term $$\rho_{3,4} = \pm\sqrt{\frac{a+b}{b-a}} = \pm \sqrt{1+\frac{2a}{b-a}}.$$

We now treat the case of $b>a$ which gives $|\rho_{1,2}|<1$ and $|\rho_{3,4}|>1.$ We may therefore apply the Cauchy Residue Theorem, taking into account the contributions from $\rho_{1,2}$ and from $z=0.$

The residues at $\rho_{1,2}$ are $$\left. \frac{(z^2+1)^2} {a^2\times 2(z^2+1)2z-b^2\times 2(z^2-1)2z} \frac{1}{iz}\right|_{z=\rho_{1,2}.}$$

This is $$\frac{1}{4i} \left. \frac{(z^2+1)^2} {a^2\times (z^2+1)z^2-b^2\times (z^2-1)z^2} \right|_{z=\rho_{1,2}} \\ = \frac{1}{4i} \left. \frac{(z^2+1)^2} {a^2\times (z^2+1)^2-b^2\times (z^2-1)^2 - a^2(z^2+1) - b^2(z^2-1)} \right|_{z=\rho_{1,2}.}$$

Considering the definition of $\rho_{1,2}$ this becomes $$-\frac{1}{4i} \left. \frac{(z^2+1)^2} {a^2(z^2+1) + b^2(z^2-1)} \right|_{z=\rho_{1,2}}$$

which yields $$-\frac{1}{4i} \frac{4b^2} {a^2\times 2b(a+b) + b^2\times (-2a)(a+b)} \\ = -\frac{1}{i} \frac{b^2} {2ab(a-b)(a+b)} = -\frac{1}{2i} \frac{b} {a(a-b)(a+b)}.$$

The residue at $z=0$ is $$\frac{1}{i} \frac{1}{a^2-b^2} = \frac{1}{i} \frac{1}{(a-b)(a+b)}.$$

Collecting everything we obtain $$\frac{1}{2}\times 2\pi i\times \left(\frac{1}{i} \frac{1}{(a-b)(a+b)} - 2\times \frac{1}{2i} \frac{b}{a(a-b)(a+b)}\right) \\ = \frac{1}{2}\times 2\pi i\times \frac{1}{i} \frac{1}{(a-b)(a+b)} \left(1 - \frac{b}{a}\right) \\ = \frac{1}{2}\times 2\pi i\times \frac{1}{i} \frac{1}{(a-b)(a+b)} \frac{a-b}{a} \\ = \frac{\pi}{a(a+b)}.$$

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