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$$ \int_0^{2 \pi} \frac{\sin ⁡x}{(1+\frac{1}{2}\cos x)} dx $$ I want to solve this.

I have solved this here in the image but I have not reached a real number

Matti P.
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Nima_Ebr
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  • Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Jun 15 '20 at 09:02
  • There's absoluteny no chance of understanding anything in the picture. Please use MathJax to write the equations. – Matti P. Jun 15 '20 at 09:12
  • Is this the integral that you want to calculate? $$ \int_0^{2 \pi} \frac{\sin \theta}{1+\frac{\cos \theta}{2}}, d\theta $$ – Matti P. Jun 15 '20 at 09:13
  • yes sir i would appreciate it if you solve it tnx. – Nima_Ebr Jun 15 '20 at 10:32
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    Perhaps you can see the integrand in the form $g'(x) \cdot f'(g(x))$ ? (at least if you multiply the integrand by $-\frac{1}{2}$) – Matti P. Jun 15 '20 at 11:08
  • i want to solve it with sinx=1/2(z+1/z). could you please guide me with this? – Nima_Ebr Jun 15 '20 at 13:05

1 Answers1

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Recall $\sin(\theta)=\frac{e^{i\theta}-e^{-i\theta}}{2i}$ and $\cos(\theta)=\frac{e^{i\theta}+e^{-i\theta}}{2}$. Use the substitution $z= e^{i\theta} $ then, $dz = i e^{i\theta} d\theta$.

The integral becomes:

$$ \frac{2}i\oint_C \frac{(z-1/z)dz}{4+z+1/z} = \frac{2}i\oint_C \frac{(z^2 -1)dz}{z^2 + 4z +1} $$ where $C$ is the unit circle around the origin. The poles of the denominator appear to be at $z_{\pm} = -2 \pm \sqrt{3} $. Hence, we see that our contour includes only $z_{+}$.

$Res(z_{+})=-2 + \sqrt{3}$. This implies $I=4 \pi (-2 + \sqrt{3})$. We need the imaginer part of this result since the initial integral was containing $sin(x)=Im[e^{i \theta}].$ Finally,

$\int_0^{2 \pi} \frac{\sin ⁡x}{(1+\frac{1}{2}\cos x)} dx =0.$

sakamoto
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