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:
$\displaystyle\frac{2}{3}\tan^{-1}\left(\frac{\tan(\frac{x}{2})}{3}\right)$
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?
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?
you have done the intregration corretly . It will have multiple answers!
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.
