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Evaluate the integral

$\int_0^{\pi}\frac{x}{a^2*\cos^2(x) + b^2*\sin^2(x)}\; dx$

In my textbook solution the integral has been split into intervals from $0$ to $\pi/4$ and then from $\pi/4$ to $\pi/2$. My question is what is the need to this? Why can't we compute the integral directly with $u=\tan x$ substitution? Any help would be appreciated.

hardmath
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user34304
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    After $;14;$ months being a member and $; 92;$ questions asked, I think it is about time you learn the easy directions to properly write mahtematics in this site: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-qu%E2%80%8C%E2%80%8Bick-reference – Timbuc Mar 16 '15 at 14:19
  • It is actually useful for outside math.stackexchange too. $\LaTeX$ and $\TeX$ are widely used to write math research papers. – Pedro Mar 16 '15 at 14:35
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    Try to at least see how people changed your previous posts, watch what is between the $ signs. Look that we changed pi by \pi for instance, see that an integral is just \int_{from}^{to} and that a fraction is just \frac{nominator}{denominator}. Also see that I added a \ before cos and sin, which gave \cos and \sin. This was something that wasn't needed, but sin and cos render more nicely in my opinion when you add , but this might be personal preference. Watch the difference between $sin$ which is wihtout a \ and $\sin$ which is with a . – Pedro Mar 16 '15 at 14:56

1 Answers1

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$\tan{x}$ is not continuous on the interval of integration, since $\tan{\frac{1}{2}\pi}$ diverges. Therefore you can't use the usual substitution rule. (Notice the condition that $\varphi$ is continuous: you need to be able to take its derivative.) If you split the interval so that $\varphi$ is single-valued and continuous, then it works.

Chappers
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