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So, my problem is $$\int_{0}^{2\pi}\frac{1}{5-cosx}$$

If you were to substitute $t=\tan{\frac{x}{2}}$ it would equal to:

$\int_{0}^{0}\frac{1}{5-4\cdot\frac{1-t^2}{1+t^2}}\cdot\frac{2dt}{1+t^2} = \int_{0}^{0}\frac{2dt}{9t^2+1}= \frac{2}{3}\int_{0}^{0}\frac{3dt}{(3t)^2+1}=\frac{2}{3}\cdot(\arctan{3t})|_0^{0} = 0$

Sorry if I made calculation errors, integral-calculator estimates this integral to be $\frac{2\pi}{3}$ and I'm not sure what I've done wrong here.

That's just an example really, I've read some other questions here on stackexchange but I'm still having issues with substituting things that change boundaries to 0, because it doesn't really make any sense to me. I understand why the boundaries get changed and everything, but when the both suddenly amount to zero and the integral is definitely not zero I don't really know what to do..

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    The substitution $t=\tan (x/2)$ that you are using, is not defined at $x=\pi$ which is a point in the interval $[0,2\pi]$, hence the discrepancy. – Anurag A Jun 20 '17 at 19:06
  • I don't believe the integral should be $2\pi/3$. – Gregory Jun 20 '17 at 19:15
  • When you make your substitution $u = f(x)$ or $x = g(t)$, the functions must be $1-1$ and continuous over the region of integration. – Doug M Jun 20 '17 at 19:16
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    You probably mean $$ \int_{0}^{2\pi}\frac{dx}{5-4\cos x}=\frac{2\pi}3\neq\frac{\pi }{\sqrt{6}}=\int_{0}^{2\pi}\frac{dx}{5-\cos x}. $$ – Olivier Oloa Jun 20 '17 at 19:18
  • @DougM, actually, injectivity of the substitution over the region of integration is not necessary to use substitution. – Bob Krueger Jun 20 '17 at 19:24
  • Related (maybe duplicate?): https://math.stackexchange.com/questions/1942983/find-int-02-pi-frac15-4-cos-x-dx – Hans Lundmark Jun 20 '17 at 19:39
  • Here are some other ones where similar things happen: https://math.stackexchange.com/questions/1296351/calculating-int-0-pi-frac1ab-sin2x-dx, https://math.stackexchange.com/questions/1697352/fundamental-theorem-of-calculus-confusion-regarding-atan – Hans Lundmark Jun 20 '17 at 19:40

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As noted by @AnuragA, the substitution $t = \tan(x/2)$ is not defined/has a BIG discontinuity at $x = \pi$. To alleviate this problem but still use the same substitution, try splitting up your range of integration into two ranges of integration, inside of which your substitution is defined. That is, try $$ \int_0^{2\pi} \frac{1}{5-\cos(x)} dx = \int_0^{\pi} \frac{1}{5-\cos(x)} dx + \int_{\pi}^{2\pi} \frac{1}{5-\cos(x)} dx $$ and evaluate both integrals separately. Note that inside the intervals $(0, \pi)$ and $(\pi, 2\pi)$, your substitution $t=\tan(x/2)$ is defined (and continuous, etc.), but inside of $(0, 2\pi)$, the substitution is not always defined.

Bob Krueger
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