Finding value of $\int^{2\pi}_{0}\frac{\sin^2(x)}{a-b\cos x}dx$ without contour Integration

Question

Finding value of $\displaystyle \int^{2\pi}_{0}\frac{\sin^2(x)}{a-b\cos x}dx\;\;, a>b>0$ without contour Integration

Try: Let $\displaystyle I =\int^{2\pi}_{0}\frac{\sin^2 x}{a-b\cos x}dx=2\int^{\pi}_{0}\frac{\sin^2 x}{a-b\cos x}dx$

Using Integration by parts

$$ I =\frac{2}{b}\bigg[\sin (x)\cdot \ln|a-b\cos x|\bigg]\bigg|^{\pi}_{0}-\frac{2}{b}\int^{\pi}_{0}\cos x\cdot \ln|a-b\cos x|dx$$

$$I=-\frac{2}{b}\int^{\pi}_{0}\cos x\cdot \ln(a-b\cos x)dx$$

Could some help me to solve it, Thanks in advance

Answer 1

Your integral is equal to $$2\int_{0}^{\pi}\frac{\sin^2x}{a-b\cos x} \, dx$$ and we can express the integrand as $$\frac{\cos x} {b} +\frac{a}{b^2}+\frac{b^2-a^2}{b^2(a-b\cos x)} $$ and hence the integral in question is $$2\cdot 0+\frac{2\pi a} {b^2}+\frac{2(b^2-a^2)}{b^2}\int_{0}^{\pi}\frac{dx}{a-b\cos x} $$ The last integral can be evaluated using a tricky substitution $$(a-b\cos x) (a+b\cos y) =a^2-b^2$$ to get $\pi/\sqrt{a^2-b^2}$ and hence the desired answer is $$\frac{2\pi a} {b^2}-\frac{2\pi\sqrt{a^2-b^2}}{b^2}=\frac{2\pi}{a+\sqrt{a^2-b^2}}$$

Answer 2

$b<a$ or the integral will not converge.

If you don't want to use complex analysis, Try the substitution:

$t = \tan \frac {x}{2}\\ x = 2\arctan t\\ dx = \frac {2}{1+t^2}\ dt\\ \sin x = \frac {2t}{1+t^2}\\ \cos x = \frac {1-t^2}{1+t^2}$

Due to the discontinutity of the $\tan$ function, this will work better if we integrate from $-\pi$ to $\pi.$ And since this is a still a full period for the integrand that is okay.

The limits of integration will be $(-\infty, infty)

$\int_{-\infty}^{\infty} \frac {8t^2}{(t^2+1)^2(a(1+t^2) + b(1-t^2))}\ dt\\ \int_{-\infty}^{\infty} \frac {8t^2}{(t^2+1)^2( (a+b) + (b-a)t^2)}\ dt$

Which looks like it is asking for a partial fraction decomposition.

$\frac {At + B}{(t^2 + 1)} + \frac {Ct + D}{(t^2 + 1)^2} + \frac {Et+F}{(a+b)t^2 + (b-a)}$

I get:

$\frac {2(b-a)}{a^2} \frac {1}{(t^2 + 1)} + \frac {4}{a} \frac {1}{(t^2 + 1)^2} + \frac {2(b^2-a^2)}{a^2} \frac {1}{(a+b)t^2 + (b-a)}$

$\frac {2(b-a)}{a^2} \arctan {t} + \frac {2}{a} (\frac {t}{1+t^2} + \arctan t) + \frac {2\sqrt {b^2-a^2}}{a^2}\arctan \sqrt {\frac {b-a}{a+b}} t$

$(\frac {(b-a)}{a^2} + \frac {1}{a}+ \frac {\sqrt {b^2-a^2}}{a^2}) 2\pi$

And double check my algebra, because I could well have dropped something along the way.

Answer 3

From $a>b$ then $\left|\dfrac{b}{a}\cos x\right|<1$ with the expansion \begin{align} \int_{0}^{2\pi}\frac{\sin^2(x)}{a-b\cos x}\ dx &= \dfrac{1}{a}\int_{0}^{2\pi}\sin^2(x)\sum_{n\geq0}\left(\dfrac{b}{a}\cos x\right)^n\ dx \\ &= \dfrac{1}{a}\sum_{n\geq0}\left(\dfrac{b}{a}\right)^n\int_{0}^{2\pi}\sin^2x\cos^nx\ dx \\ &= \dfrac{1}{a}\sum_{n\geq0, \text{even}}\left(\dfrac{b}{a}\right)^n\left(\dfrac{2\pi}{2^n}{n\choose n/2} - \dfrac{2\pi}{2^{n+2}}{n+2\choose n/2+1}\right) \\ &= \color{blue}{\dfrac{2\pi}{a+\sqrt{a^2-b^2}}} \end{align} Using here $$\int_0^{2\pi}\cos^n x\ dx=\dfrac{2\pi}{2^n}{n\choose n/2}$$ for even $n$ and for odd $n$ it is zero.

Answer 4

$$I=\int_0^{2\pi}\frac{\sin^2(x)}{a-b\cos(x)}dx=\int_0^\pi\frac{\sin^2(x)}{a-b\cos(x)}dx+\int_\pi^{2\pi}\frac{\sin^2(x)}{a-b\cos(x)}dx$$ now using substitution: $$t=\tan\frac{x}{2}$$ $$dx=\frac{2}{1+t^2}dt$$ $$\sin(x)=\frac{2t}{1+t^2}$$ $$\cos(x)=\frac{1-t^2}{1+t^2}$$ because of the discontinuities of the function, we must first change the limits. Now our integral becomes: $$I=\int_{-\pi}^\pi\frac{\sin^2(x)}{a-b\cos(x)}dx=\int_{-\infty}^\infty\frac{\left(\frac{2t}{1+t^2}\right)^2}{a-b\left(\frac{1-t^2}{1+t^2}\right)}.\frac{2}{1+t^2}dt=2\int_{-\infty}^\infty\frac{2t}{(1+t^2)^2\left(a(1+t^2)-b(1-t^2)\right)}dt=2\int_{-\infty}^\infty\frac{2t}{(1+t^2)^2\left((a-b)+(a+b)t^2\right)}dt$$ then by using PFD this can be evaluated