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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)

Zev Chonoles
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Ashish Gaurav
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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

2 Answers2

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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*}

Sangchul Lee
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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. – Sangchul Lee Feb 23 '13 at 17:48
  • I think you also used that $$\int_0^\pi xf(\sin x)dx=\frac \pi 2 \int_0^\pi f(\sin x)dx$$ – Pedro Feb 23 '13 at 17:59
  • @PeterTamaroff: That's also right, though I skipped the explanation because OP was already well aware of that. – Sangchul Lee Feb 23 '13 at 18:25
  • @SangchulLee Am I missing something here or is $b \tan x \mapsto a \tan x$ a substitution without any change of variable? $b \tan x \mapsto a \tan u$, for example, is what I'm used to seeing. – s0ulr3aper07 Mar 01 '19 at 01:54
  • @s0ulr3aper07, I would not say this is a good practice, but I overrode the meaning of the variable $x$ to a new one. The main purpose of doing so, even at the risk of introducing possible source of ambiguity, is not to introduce many variable names, especially when they are only dummy variables. – Sangchul Lee Mar 01 '19 at 16:23
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$$ \begin{aligned} \because I&=\int_0^\pi \frac{x}{\left(a^2 \cos ^2 x+b^2 \sin ^2 x\right)^2} \\& \stackrel{x\mapsto\pi-x}{=} \int_0^\pi \frac{\pi-x}{\left(a^2 \cos ^2 x+b^2 \sin ^2 x\right)^2} d x \\ &=\pi \int_0^\pi \frac{d x}{\left(a^2 \cos ^2 x+b^2 \sin ^2 x\right)^2}-I \\ \therefore I&=\frac{\pi}{2} \int_0^\pi \frac{d x}{\left(a^2 \cos ^2 x+b^2 \sin ^2 x\right)^2} \\ &=\pi \int_0^{\frac{\pi}{2}} \frac{d x}{\left(a^2 \cos ^2 x+b^2 \sin ^2 x\right)^2} \end{aligned} $$

By my post, $$ \boxed{I =\pi\left[\frac{\pi\left(a^2+b^2\right)}{4 a^3 b^3}\right]=\frac{\pi^2\left(a^2+b^2\right)}{4 a^3 b^3}} $$

Lai
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