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$$\int_0^{2\pi} \frac{1}{2+\sin(x)} \ dx$$

$$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$

$$z=\alpha(x)=e^{ix}$$

$$\sin(x)=\frac{z^2-1}{2zi}$$ $$2+\sin(x)=\frac{4zi+z^2-1}{2zi}$$ $$\frac{1}{2+\sin(x)}=\frac{2zi}{4zi+z^2-1}$$


$$\int_\alpha \frac{2zi}{4zi+z^2-1} \ \ \frac{1}{zi}$$


$$4zi+z^2-1=0$$ $$z_1=\frac{-4i+2\sqrt{3}i}{2}$$ $$z_2=\frac{-4i-2\sqrt{3}i}{2}$$


$$\rvert z_1 \rvert <1$$ $$\rvert z_2 \rvert >1$$


Using Residue theorem:

$$\int_\alpha \frac{2}{4zi+z^2-1}=2 \pi i \ \lim_{z\rightarrow-2i+\sqrt{3}i} \ \frac{2}{z+2i+\sqrt{3}i}=\frac{2 \pi}{\sqrt{3}}$$


Is it correct?

Thanks!

  • 3
    Yes your work seems to be fine!! – 2017-01-10
  • 0
    The mathematics is correct, but you are missing a few $dz$'s in your integrals, and writing a bit more text on what you're doing between each step would certainly be helpful! – 2017-01-10
  • 0
    you can always check your work by using some CAS like WolframAlpha... – 2017-01-10

3 Answers 3

2

Yes, absolutely correct but you may try alternative approach by converting $\sin\space x$ into $\tan\space (x/2)$ $$\sin(x) = \frac{2\cdot \tan(x/2)}{1+\tan^2(x/2)}$$.

1

Another way is

Set $u=tg\frac{x}{2}$ and $du=sec^2\frac{x}{2}*\frac{1}{2}dx \rightarrow dx=\frac{2}{1+u^2}dt$ and $sin(x)=\frac{2u}{u^2+1}$

$$\int_0^{2\pi} \frac{1}{2+\sin(x)} \ dx=\int_{-\pi}^{\pi} \frac{1}{2+\sin(x)} \ dx= \int_{-\infty}^{\infty} \frac{2}{(u^2+1)(\frac{2u}{u^2+1}+2)}=\int_{-\infty}^{\infty} \frac{1}{u^2+u+1}$$

And to solve this just complete the square and you will get something in function of $arctg(u)$

  • 0
    It might be worth mentioning that one should shift the interval of integration to $(-\pi,\pi)$ before doing the substitution. – 2017-01-10
1

\begin{align} \int_0^{2\pi} \frac{1}{2+\sin(x)} dx&=\int_0^{2\pi} \frac{1}{1+\sin^2(x/2)+\cos^2(x/2)+2\sin(x/2)\cos(x/2)} dx\\ &=\int_0^{2\pi} \frac{1}{1+(\sin(x/2)+\cos(x/2))^2} dx\\ &=\int_0^{2\pi} \frac{1}{1+2\sin^2(x/2+\pi/4)} dx\\ \end{align}

From here you should be able to continue recalling the behaviour of $\arctan$.