# Contour integration problem with sin and cos

so I'm revising contour integration for an upcoming complex analysis exam. I have been asked to integrate $$\int_0^{2\pi}\frac{\sin^2x}{a+b \cos x}dx$$

I thought the sensible thing to do here would be to substitute in $$z=e^{ix}$$ and take a contour integral around the unit circle, call this path $$\ast$$ so that my integral becomes $$\frac{1}{2i}Re(\int_\ast\frac{1-z^2}{az+bz^2}dz)$$

Then, letting $$f(z)=\frac{1-z^2}{az+bz^2}$$, I thought the function had simple poles at $$z=0$$ with residue $$\frac{1}{a}$$ and another simple pole at $$z=\frac{-a}{b}$$ with residue $$\frac{a}{b^2}-\frac{1}{a}$$ and thus I get the that $$Re(\int_\ast\frac{1-z^2}{az+bz^2}dz)=2i(2\pi i)\frac{a}{b^2}=-4\pi(\frac{a}{b^2})$$ which is not the answer given, that is: $$=\frac{2\pi}{b^2}[a-\sqrt{a^2-b^2}]$$,but I can't work out why.

Any help appreciated, thank you in advance.

• I got $$\frac{-i}{2}\int_*\frac{(1-z^2)^2}{z^2(bz^2+2az+b)}\;dz$$ May 22 '19 at 17:30
• You mistake is that $Re(x)/Re(y)\neq Re(x/y)$ May 22 '19 at 17:37
• Is there some assumption about the values of $a$ and $b$? May 22 '19 at 17:40
• Ah I see, thank you so much. Assumption on a and b is that a>b>0. May 22 '19 at 17:51

\begin{align} \int_0^{2\pi}\frac{\color{#C00}{\sin^2(x)}}{\color{#090}{a+b\cos(x)}}\,\color{#00F}{\mathrm{d}x} &=\oint\frac{\color{#C00}{-\frac{z^4-2z^2+1}{4z^2}}}{\color{#090}{\frac{bz^2+2az+b}{2z}}}\color{#00F}{\frac{\mathrm{d}z}{iz}}\\ &=-\frac1{2i}\oint\frac{z^4-2z^2+1}{bz^2+2az+b}\,\frac{\mathrm{d}z}{z^2} \end{align} The residue of $$\frac{z^4-2z^2+1}{bz^2+2az+b}\frac1{z^2}$$ at $$z=\frac{-a\pm\sqrt{a^2-b^2}}b$$ (simple poles) is $$\pm\frac{2\sqrt{a^2-b^2}}{b^2}$$ and the residue at $$z=0$$ (degree $$2$$ pole) is $$-\frac{2a}{b^2}$$. Assuming $$a\gt0$$, we get $$2\pi i$$ times the sum of the residues inside the unit circle, $$\frac{-a+\sqrt{a^2-b^2}}b$$ and $$0$$, to be $$\int_0^{2\pi}\frac{\sin^2(x)}{a+b\cos(x)}\,\mathrm{d}x=2\pi\frac{a-\sqrt{a^2-b^2}}{b^2}$$

If you substitute $$z = e^{ix}$$, then $$\sin(x) = (z - \frac{1}{z})/(2i)$$ and $$\sin^2(x) =(z-\frac{1}{z})^2/(-4) = \frac{(z^2-1)^2}{-4z^2}$$

$$dz = ie^{ix}dx = izdx$$ or $$dx = \frac{dz}{zi}$$

$$a+b\cos x = a + b\frac{z+\frac{1}{z}}{2} = \frac{2az + bz^2 +1}{2z}$$

So,

$$I = \int_*\frac{\frac{(z^2-1)^2}{-4z^2}dz}{zi\frac{2az + bz^2 +1}{2z}}$$

$$I = \int_*\frac{i(z^2-1)^2dz}{2z^2(2az + bz^2 +1)}$$

• Quickly typed, but please check sign and that one in the denominator, then how the $z$-s cancel in the denominator of the first $I$, and the $z^{-2}$ from the numerator survives... May 22 '19 at 17:38
– Ak.
May 22 '19 at 17:39

We suppose $$b\ne 0$$, and get rid of it by force, use rather instead of $$a,b$$ only the variable $$c=a/b$$.

(After some edits in the OP we have indeed $$0, thus $$1.)

We will assume $$c$$ real, $$|c|>1$$, so that there is no zero in the denominator. (And the integral does not diverge.)

Sooner or later we will have to fight against the two roots of $$z^2+2c+1$$, we denote them by $$U,V$$, their product is $$UV=1$$, and thus we may and do assume $$|U|<1$$, $$|V|>1$$.

Let $$C$$ be the unit circle centered in the origin of the complex plane, then formally using $$z=e^{ix}$$, $$\frac 1{iz}\; dz=dx$$ we have: \begin{aligned} J &= \int_0^{2\pi}\frac{\sin^2x}{a+b\cos x}\;dx \\ &= \frac 1b\int_0^{2\pi}\frac{\sin^2x}{\cos x+c}\;dx \\ &= \frac 1b \int_C \frac{ \left(\frac 1{2i}\left(z-\frac 1z\right)\right)^2} {\frac 12\left(z+\frac 1z\right)+c}\;\frac 1{iz}\;dz \\ &= \frac i{2b} \int_C \frac{ (z^2-1)^2} {z^2(z^2+2cz+1)}\;dz\ . \\[3mm] &\qquad\text{ The partial fraction decomposition is:} \\ \frac{ (z^2-1)^2} {z^2(z^2+2cz+1)} &= \frac{ (z^2-1)^2} {z^2(z-U)(z-V)} \\ &= 1+\frac 1{z^2}+\frac{U+V}z -\frac {U-V}{z-V} -\frac {V-U}{z-U} \ . \\[3mm] &\qquad\text{ Only the residues in 0,U contribute, so...} \\ J&=2\pi i\cdot \frac i{2b} \cdot[(U+V)+(U-V)] \\ &=2\pi \frac {-U}b =2\pi \frac {c-\sqrt{c^2-1}}b \ . \end{aligned} The last expression corresponds to the expected answer from the OP, recalling that $$c=a/b$$.

Sage check for the partial fraction decomposition:

sage: var('U,z');
sage: V = 1/U
sage: EE = (z^2-1)^2 / z^2 / (z-U) / (z-V)
sage: EE.partial_fraction(z)
-(U^2 - 1)/(U*z - 1) - (U^2 - 1)/((U - z)*U) + (U^2 + 1)/(U*z) + 1/z^2 + 1
sage: bool( _ == (1 + 1/z^2 + (U+V)/z - (V-U)/(z-U) - (U-V)/(z-V) ) )
True