Question

Solve:
${\int }_0^{\pi /2}\frac{1+2\cos x}{(2+\cos x{)}^2}dx$

Answer 1

$I={\int }_0^{\pi /2}\frac{1+2\cos x}{(2+\cos x{)}^2}dx={\int }_0^{\pi /2}\frac{2(\cos x+2)-3}{(2+\cos x{)}^2}dx$

$={\int }_0^{\pi /2}\frac{x}{\cos x+2}dx-3{\int }_0^{\pi /2}\frac{dx}{(2+\cos x{)}^2}dx$

$I_1={\int }_0^{\pi /2}\frac{2}{\cos x+2}dx={\int }_0^{\pi /2}\frac{2}{\frac{1+{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}}={\int }_0^{\pi /2}\frac{2{\sec }^2\frac{x}{2}dx}{{\tan }^2\frac{x}{2}+3}$

Let $u=\tan \frac{\left(x\right)}{2}\Rightarrow du=\frac{1}{2}{\sec }^2\frac{x}{2}dx$

When $x=0,u=0$ & $x=\frac{\pi }{2},u=1$

$I_1={\int }_0^1\frac{4du}{4^2+3}=\frac{4}{3}{\int }_0^1\frac{du}{{\left(\frac{u}{\sqrt{3}}\right)}^2+1}=\frac{4}{\sqrt{3}}{\left[{\tan }^{-1}\frac{4}{\sqrt{3}}\right]}_0^1=\frac{4}{\sqrt{3}}\left(\frac{\pi }{6}\right)=\frac{2\pi }{3\sqrt{3}}$ ............ $\left(1\right)$

$I_2=-3{\int }_0^{\pi /2}\frac{dx}{(\cot x+2{)}^2}=-3{\int }_0^{\pi /2}\frac{dx}{{|\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}+2|}^2}$

$=-3{\int }_0^{\pi /2}\frac{(1+{\tan }^2\frac{x}{2}){\sec }^2\frac{x}{2}}{{({\tan }^2\frac{x}{2}+3)}^2}$

Let $V=\tan \frac{x}{2}\Rightarrow dv=\frac{1}{2}{\sec }^2\frac{x}{2}dx$

when $x=0,v=0$ & $x=\frac{\pi }{2},v=1$

$I_2=-3{\int }_0^1\frac{1+v^2}{(3+v^2{)}^2}2dv=-6{\int }_0^1\frac{(v^2+3)-2}{(v^2+3{)}^2}dv$

$=-6\left[{\int }_0^1\frac{dv}{v^2+3}-{\int }_0^1\frac{2}{(v^2+3{)}^2}dv\right]$

$-6{\left[\frac{\sqrt{3}}{3}{\tan }^{-1}\frac{v}{\sqrt{3}}\right]}_0^1+12{\int }_0^1\frac{dv}{(v^2+3{)}^2}$

$=\frac{-6\sqrt{3}}{3}\left(\frac{\pi }{6}\right)+I_3$

$=\frac{-\pi }{\sqrt{3}}+I_3$ .............. $\left(2\right)$

$I_3=12{\int }_0^1\frac{dv}{(v^2+3{)}^2}$

Let $v=\sqrt{3}\tan \theta \Rightarrow dv=\sqrt{3}{\sec }^2\theta d\theta$

when $v=0,\theta =0$ & $v=1,\theta =\frac{\pi }{6}$

$I_3=12{\int }_0^{\pi /6}\frac{\sqrt{3}{\sec }^2\theta d\theta }{9({\tan }^2\theta +1{)}^2}=\frac{4}{\sqrt{3}}{\int }_0^{\pi /6}{\cos }^2\theta d\theta$

$=\frac{4}{\sqrt{3}}{\int }_0^{\pi /6}\left(\frac{1+\cos 2\theta }{2}\right)d\theta$

$=\frac{2}{\sqrt{3}}{[\theta +\frac{\sin 2\theta }{2}]}_0^{\pi /6}$

$=\frac{2}{\sqrt{3}}(\frac{\pi }{6}+\frac{\sqrt{3}}{4})$ ............ $\left(3\right)$

$I_2=\frac{\pi }{\sqrt{3}}+\frac{2}{\sqrt{3}}\left(\frac{\pi }{6}\right)+\frac{1}{2}=\frac{-2\pi }{3\sqrt{3}}+\frac{1}{2}$ ........ $\left(4\right)$

from $\left(1\right),\left(4\right)$ $I=I_1+I_2=\frac{2\pi }{3\sqrt{3}}+(\frac{1}{2}-\frac{2\pi }{3\sqrt{3}})$

$I=\frac{1}{2}$