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May I please get help to solve the following expression, $$\frac 1 {2\pi}\int \frac 1 {(1-\alpha \cos \omega)^2 +\alpha^2\sin^2 \omega} \, d\omega$$

I have tried several ways, but couldn't get through. Thanks in advance.

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1 Answer 1

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Simply put $\tan\frac{\omega}{2}=t $ to get

$$\frac{1}{2\pi}\int\frac{dt}{(a^2+1)(t^2+1)-2a(1-t^2)}$$

The answer is therefore

$$\frac{1}{2\pi|a^2-1|}\arctan\left( \frac{|a+1|t}{|a-1|}\right)+c$$

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  • $\begingroup$ Thank you for your reply. Pardon my ignorance, can you please elaborate the steps in between. Thanks in advance. $\endgroup$ Feb 4, 2018 at 14:57
  • $\begingroup$ First use that $\cos^2(x)+\sin^2(x)=1$. now when doing that substitution, you need to convert $\cos\omega$ in terms of t. That is given by $\cos(2u)=\frac{1-\tan^2(u)}{1+\tan^2(u)}$ . $\endgroup$
    – King Tut
    Feb 4, 2018 at 15:05
  • $\begingroup$ After that use standard integral $\int \frac{dx}{a^2+x^2}$.. $\endgroup$
    – King Tut
    Feb 4, 2018 at 15:07
  • $\begingroup$ Thank you so much! It really helped me. $\endgroup$ Feb 4, 2018 at 15:11
  • $\begingroup$ You are welcome! $\endgroup$
    – King Tut
    Feb 4, 2018 at 15:18

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