Question

Evaluate ${\int }_0^{\frac{\pi }{2}}\frac{dx}{3+2\cos x}.$

Answer 1

$I={\int }_0^{\pi /2}\frac{dx}{3+2cosx}={\int }_0^{\pi /2}\frac{dx}{3+2\left(\frac{1-ta{n}^{2}x/2}{1+ta{n}^{2}x/2}\right)}$

$={\int }_0^{\pi /2}\frac{se{c}^{2}x/2dx}{5+ta{n}^{2}x/2}$

Let $z=tanx/2$

$\Rightarrow dz=\frac{1}{2}se{c}^{2}x/2dx.$

$\Rightarrow 2dz=se{c}^{2}x/2dx$

Also limit of $z=0toz=1$

$I={\int }_0^1\frac{2dz}{5+z^2}=2{\int }_0^1\frac{dz}{z^2+(\sqrt{5}{)}^2}$

$=\frac{2}{\sqrt{5}}{\left[ta{n}^{-1}\left(\frac{z}{\sqrt{5}}\right)\right]}_0^1$

$=\frac{2}{\sqrt{5}}ta{n}^{-1}\left(\frac{1}{\sqrt{5}}\right)$

$\therefore {\int }_0^{\pi /2}\frac{dx}{3+2cosx}=\frac{2}{\sqrt{5}}ta{n}^{-1}\left(\frac{1}{\sqrt{5}}\right).$