# Evaluating $\int_{0}^{2 \pi} \frac{\sin(\phi)}{(\sin(\theta)-\sin(\phi))^2+2a(1-\cos(\theta -\phi))} d\phi$

How Can I evaluate this integral ? $$\int_{0}^{2 \pi} \frac{\sin(\phi)}{(\sin(\theta)-\sin(\phi))^2+2a(1-\cos(\theta -\phi))} d\phi$$

I have tried to write the denominator as $$\sin( (\theta-\phi)/2 )[2\cos((\theta+\phi)/2)+4a\sin((\theta-\phi)/2)]$$

And writing the numerator as $$\sin(\phi)=2\sin((\phi-\theta)/2)\cos((\phi - \theta)/2)\cos(\theta)+\sin(\theta)(1-2\sin^2((\phi-\theta)/2)$$ Where $$\theta , \phi \in[0,2 \pi]$$

• Did you try to work it out using a complex curve integral? I ask, since you use those tags. If so, where did you get stuck in that attempt? – mickep May 6 at 19:45
• Yes I did try using $z=e^{i \phi}$ but I didn't get the result . – TopSpin May 11 at 16:04

Here $$\theta=t$$, $$\phi=x$$. Your integral is $$q(a,t)=\int_0^{2\pi}\frac{\sin x}{(\sin t-\sin x)^2+2a(1-\cos(t-x))}dx$$ Note that $$\sin\alpha-\sin\beta=2\sin\left(\frac{\alpha-\beta}2\right)\cos\left(\frac{\alpha+\beta}2\right)$$ and $$1-\cos(\alpha-\beta)=2\sin^2\left(\frac{\alpha-\beta}2\right)$$ So we have \begin{align} q(a,t)=&\frac14\int_0^{2\pi}\frac{\sin x}{\sin^2(\frac{t-x}2)(2a+\cos^2(\frac{t-x}2))}dx\\ =&\frac14\int_0^{2\pi}\frac{\sin x}{\sin^2(\frac{x}2-\frac{t}2)\cos^2(\frac{x}2-\frac{t}2)+2a\sin^2(\frac{x}2-\frac{t}2)}dx\\ \overset{2u=x-t}=&\frac12\int_{-t/2}^{\pi-t/2}\frac{\sin(2u+t)}{\sin^2(u)\cos^2(u)+2a\sin^2(u)}du\\ =&\frac12\int_{-t/2}^{\pi-t/2}\frac{c\sin(2u)+s\cos(2u)}{\cos^2(u)\sin^2(u)+2a\sin^2(u)}du\\ =&-\frac12\int_{-t/2}^{\pi-t/2}\sec^2(u)\frac{s\tan^2(u)-2c\tan(u)-s}{\tan^2(u)(2a\tan^2(u)+2a+1)}du\\ \overset{x=\tan(u)}=&-\frac12\int_{x_1}^{x_2}\frac{sx^2-2cx-s}{x^2(2ax^2+2a+1)}dx \end{align} Where $$s=\sin(t)$$, $$c=\cos(t)$$, $$x_1=\arctan(-t/2)$$, and $$x_2=\arctan(\pi-t/2)$$. At this point we preform partial fractions to get $$q(a,t)=-\frac1{2(2a+1)}\left[\int_{x_1}^{x_2}\frac{4acx+(4a+1)s}{2ax^2+2a+1}dx-2c\int_{x_1}^{x_2}\frac{dx}{x}-s\int_{x_1}^{x_2}\frac{dx}{x^2}\right]$$ Can you take it from here?
• Nice! I got to the $2u=x-t$ part then got lost in the many terms. This is a cool way of handling them. – user1952500 May 7 at 3:33
• @user1952500 I'm glad you liked it! It is always helpful (for me at least) to represent constants (like $\sin t$ or $\cos t$ in this case) by single letters, so I don't get confused by what contains an $x$ and what doesn't :) – clathratus May 7 at 3:36
• $\cos^2(x/2 -t/2)$ or $\cos^2(x/2+t/2)$ – TopSpin May 11 at 15:10