Question

Evaluate ${\int }_0^{2\pi }\frac{{\sin }^2\theta }{a-b\cos \theta }d\theta$ for $a>b>0$

• A. $\frac{2\pi }{b^2}[a-\sqrt{(a^2-b^2)}].$
• B. $\frac{2\pi }{a^2}[-a+\sqrt{(a^2+b^2)}].$
• C. $\frac{2\pi }{a^2}[a+\sqrt{(a^2-b^2)}].$
• D. $\frac{2\pi }{b^2}[a+\sqrt{(a^2+b^2)}].$

Answer 1

The correct option is A $\frac{2\pi }{b^2}[a-\sqrt{(a^2-b^2)}].$

Let $I={\int }_0^{2\pi }\frac{{\sin }^2\theta d\theta }{a-b\cos \theta }=2{\int }_0^{\pi }\frac{{\sin }^2\theta d\theta }{a-b\cos \theta }$

Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

$2I=2{\int }_0^{\pi }\frac{2a{\sin }^2\theta }{a^2-b^2{\cos }^2\theta }d\theta =8a{\int }_0^{\pi /2}\frac{{\sin }^2\theta d\theta }{a^2-b^2{\cos }^2\theta }$

Dividing numerator and denominator by ${\cos }^2\theta$

$2I=8a{\int }_0^{\pi /2}\frac{{\tan }^2\theta d\theta }{a^2(1+{\tan }^2\theta )-b^2}=8a{\int }_{\pi /2}^0\frac{{\tan }^2\theta d\theta }{(a^2-b^2+a^2{\tan }^2\theta )}$

Put $a\tan \theta =t$

$\therefore I=4{\int }_0^{\infty }\frac{\left(t^2/a^2\right)a^2}{[(a^2-b^2)+t^2][a^2+t^2]}=4{\int }_0^{\infty }[\frac{a^2}{b^2(a^2+t^2)}-\frac{a^2-b^2}{b^2\{(a^2-b^2)+t^2\}}]dt$

$=\frac{4}{b^2}{[a^2,\frac{1}{a}{\tan }^{-1}\frac{t}{a}-\frac{(a^2-b^2)}{\sqrt{(a^2-b^2)}}{\tan }^{-1}\frac{t}{\sqrt{(a^2-b^2)}}]}_0^{\infty }$

$=\frac{4}{b^2}[a.\frac{\pi }{2}-\sqrt{(a^2-b^2).\frac{\pi }{2}}]=\frac{2\pi }{b^2}[a-\sqrt{(a^2-b^2)}]$

$\therefore I=\frac{2\pi }{b^2}[a-\sqrt{(a^2-b^2)}]$