We are doing definite integrals in university and I wanted to practice but this problem is giving me a hard time.

The problem is to evaluate the following integral:

$ \displaystyle \int_{0}^{2\pi} \frac{1}{5+4\cos(x)} dx$

For the antiderivative I got:


Now the result should be $\frac{2\pi}{3}$, but all I get with splitting the integral is always $0$.

Can someone help me find the right way to do this calculation?

  • $\begingroup$ @ Paul H...Calculus of residue can be of use $\endgroup$ – Anupam Oct 3 '13 at 16:00
  • $\begingroup$ Your antiderivative is false. $\endgroup$ – user37238 Oct 3 '13 at 16:01
  • $\begingroup$ @user37238 why is it wrong? if i derive it, i get the integrand $\endgroup$ – PaulH Oct 3 '13 at 16:08
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    $\begingroup$ The OP says they're doing definite integrals in the university. Chances are this is a first year calculus I exercise and thus complex analysis still is two or more years away. $\endgroup$ – DonAntonio Oct 3 '13 at 16:16
  • $\begingroup$ @Anupam I googled it. We haven't learned that yet. $\endgroup$ – PaulH Oct 3 '13 at 16:18

To avoid nonsenses I'd rather go with a trigonometric (Weierstrass's) substitution:

$$t=\tan\frac x2\implies \begin{cases}\cos x=\frac{1-t^2}{1+t^2}\\{}\\dx=\frac2{1+t^2}dt\end{cases}$$

and I choose the limits of the original integral to be $\;-\pi\;,\;\pi\;$ (why is it possible?), so after the substitution we get for $\;-\infty <t<\infty\;$ , and we need to solve

$$\int\limits_{-\infty}^\infty\frac{1}{5+4\frac{1-t^2}{1+t^2}}\cdot\frac2{1+t^2}dt=4\int\limits_0^\infty\frac{dt}{9+t^2}=\frac43\int\limits_0^\infty\frac{\frac13dt}{1+\left(\frac t3 \right)^2}=$$

$$=\left.\frac43\arctan\frac t3\right|_0^\infty=\frac43\left(\frac\pi2\right)=\frac{2\pi}3$$

Note that $\;4\;$ in the middle integral above: why did I multiply the whole thing by two and changed the lower limit?


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you have done the intregration corretly . It will have multiple answers!

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    $\begingroup$ Why the plus sign intead of minus in the second and third lines? $\endgroup$ – DonAntonio Oct 3 '13 at 16:18
  • $\begingroup$ How do i know the 'right' answer? When i look at the graph, it sure is $\frac{2\pi}{3}$ $\endgroup$ – PaulH Oct 3 '13 at 16:24

U have substituted $tan(\frac{x}{2})$ which can be done had it been continuous, differentially and monotonic. Break the integral from 0 to $\pi$ and $\pi$ to $2\pi$. You will get right answer.


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