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Evaluate $\int_0^{\pi}\frac{dx}{a^2\cos^2x +b^2 \sin^2x}$

I got numerator $\sec^2 x$ and denominator $b^2 ( a^2/b^2 + \tan^2x)$. I made substitution $u= \tan x$. That way $\sec^2 x$ got cancelled and the answer was of form $1/ab$ ($\tan^{-1} (bu/a)$) And then if I put limits answer is $0$ but answer is wrong. Where did I go wrong?

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Hint: Try $~t=\pi-x,~$ simplify the integrand, then add the two equivalent integral expressions together, and use the Weierstrass substitution.


The problem lies with the fact that $\tan x$ is not bijective on $[0,\pi],~$ since $\tan0=\tan\pi=0,$ so you're ultimately evaluating $\displaystyle\int_0^0f(t)~dt=0$. My advice would be to split the integral with regard to $~\dfrac\pi2~$ first, before making any substitutions.

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  • $\begingroup$ Umm can you point out the mistake in my,process? I don't know what went wrong $\endgroup$ – Maini who Mar 17 '15 at 10:42
  • $\begingroup$ i don't think this helps.Sorry. $\endgroup$ – RE60K Mar 17 '15 at 10:57
  • $\begingroup$ @Mainiwho: Yes. I believe that the problem lies with the fact that $\tan x$ is not bijective on $[0,\pi]$, since $\tan0=\tan\pi=0$, so you're ultimately evaluating $\displaystyle\int_0^0f(t)~dt=0$. My advice would be to split the integral with regard to $\dfrac\pi2$ first, before making any substitutions. $\endgroup$ – Lucian Mar 17 '15 at 10:57
  • $\begingroup$ @ADG: What seems to be the problem ? Works perfectly fine for me. $\endgroup$ – Lucian Mar 17 '15 at 10:58
  • $\begingroup$ yes it is the first method i'd use but the op asks for his mistake and this rather would confuse me if i were her/him. $\endgroup$ – RE60K Mar 17 '15 at 11:01
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Are you talking about this?:

$$\small\int\frac{dx}{a^2\cos^2x +b^2 \sin^2x}=\int\frac{\sec^2xdx}{a^2+b^2 \tan^2x}\stackrel{u=\tan x}=\frac1{b^2}\int\frac{du}{a^2/b^2+u^2}=\frac1{b^2}\frac1{a/b}\arctan\frac{\tan x}{a/b}$$ So: $$\int_0^{\pi}\frac{dx}{a^2\cos^2x +b^2 \sin^2x}=\frac1{ab}\arctan\frac{b\tan x}a\Bigg|_0^{\pi}=0$$

This is wrong because at $\pi/2$, $\cos x=0\iff \sec x\to\infty$!! Or you can say because $\tan x$(=u) misbehaves at $x=\pi/2$.It should rather be done like:

$$\int_0^{\pi}\frac{dx}{a^2\cos^2x +b^2 \sin^2x}=\frac1{ab}\arctan\frac{b\tan x}a\Bigg|_0^{\displaystyle\pi/2^-}+\frac1{ab}\arctan\frac{b\tan x}a\Bigg|_{\displaystyle\pi/2^+}^{\displaystyle\pi}$$

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  • $\begingroup$ Yes but answer is pi^2 /2ab $\endgroup$ – Maini who Mar 17 '15 at 10:56
  • $\begingroup$ @Mainiwho see edit. $\endgroup$ – RE60K Mar 17 '15 at 10:57
  • $\begingroup$ So how do I Solve it?? $\endgroup$ – Maini who Mar 17 '15 at 10:58
  • $\begingroup$ @Mainiwho see edit. $\endgroup$ – RE60K Mar 17 '15 at 11:00
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$$\begin{eqnarray*}\color{red}{I}&=&\int_{0}^{\pi}\frac{dx}{a^2\cos^2 x+b^2\sin^2 x}=2\int_{0}^{\pi/2}\frac{dx}{a^2\cos^2 x+b^2\sin^2 x}\\&=&2\int_{0}^{+\infty}\frac{dt}{(1+t^2)\left(a^2\frac{1}{1+t^2}+b^2\frac{t^2}{t^2+1}\right)}=2\int_{0}^{+\infty}\frac{dt}{a^2+b^2 t^2}\\&=&\frac{2}{ab}\int_{0}^{+\infty}\frac{du}{1+u^2}=\color{red}{\frac{\pi}{ab}}.\end{eqnarray*}$$ Here we replaced $x$ with $\arctan t$, then $t$ with $\frac{a}{b}\,u$.

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Hint:

The problem is that you divide the numerator and the denominator of the integrand by $\cos^2 x$, but this can be done only for $\cos^2 x \ne 0$ i.e. $x\ne \pi/2$. So you have to divide your integral in two improper integrals as: $$ \int_0^\pi = \int_0^{(\pi/2)^-}-\int_\pi^{(\pi/2)^+} $$ Your substutition is correct, so you have the limit: $$ \lim_{x \rightarrow (\pi/2)^{\pm}} \arctan \left(\frac{b}{a} \tan x \right) $$ that is $\pm \pi/2$.

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