Question

Use the method of contour integration, evaluate the integral ∫02πcos3θ/5−3cosθdθ.

Answer 1

Given integral-

$I=\int_0^{2\pi} (\dfrac{1}{5}cos3\theta -3cos\theta)d\theta$

As we know, $e^{i\theta}=cos\theta+isin\theta$

and $e^{-i\theta}=cos\theta-isin\theta$

So $cos\theta=\dfrac{e^{i\theta}+e^{-i\theta}}{2}\\[9pt]sin\theta=\dfrac{e^{i\theta}-e^{-i\theta}}{2}$

Let $z=e^{i\theta}\Rightarrow dz=ie^{i\theta}d\theta$

So Our integral becomes-

$I=\int_0^{2\pi}[ \dfrac{1}{10}(e^{i3\theta}+e^{-3i\theta})-(\dfrac{e^{i\theta}-e^{-i\theta}}{2})]d\theta$

$=\int_0^1[\dfrac{1}{10}(z^3+z^{-3})-\dfrac{1}{2}(z-z^{-1})]dz\\[9pt]=\dfrac{1}{10}(\dfrac{z^4}{4}-\dfrac{2}{z^2})-\dfrac{1}{2}(\dfrac{z^2}{2}-lnz)|_0^1\\[9pt]=\dfrac{1}{10}(\dfrac{1}{4}-2)-\dfrac{1}{2}(\dfrac{1}{2}-ln1)\\[9pt]=\dfrac{-7}{40}-\dfrac{1}{4}\\[9pt]=\dfrac{-7-10}{40}=\dfrac{-17}{40}$