Question

Evaluate: ${\int }_0^{\frac{\pi }{4}}\frac{dx}{5+4cosx}$

Answer 1

${\int }_0^{\pi /4}\frac{dx}{5+4\cos x}={\int }_0^{\pi /4}\frac{dx}{5+4\left(\frac{1-{\tan }^{2}x/2}{1+{\tan }^{2}x/2}\right)}$

$={\int }_0^{\pi /4}\frac{dx(1+{\tan }^{2}x/2)}{5+5{\tan }^{2}x/2+4-4{\tan }^{2}x/2}$

$={\int }_0^{\pi /4}\frac{{\sec }^{2}x/2dx}{9+{\tan }^{2}x/2}$ Let $t=\tan x/2$ $dt=\frac{1}{2}{\sec }^{2}x/2dx$

$={\int }_0^{\pi /4}\frac{2dt}{9+t^2}$

When $x/2=0,t=0$; when $x/2=\pi /4$ $\Rightarrow t=1$

$={\int }_0^1\frac{2dt}{9+t^2}$

$={2\times \frac{1}{3}{\tan }^{-1}\left(\frac{t}{3}\right)]}_0^1$

$=\frac{2}{3}[{\tan }^{-1}\left(1/3\right)-{\tan }^{-1}(0\left)\right]$

$=\frac{2}{3}[{\tan }^{-1}\left(1/3\right)-0]$

$=\frac{2}{3}{\tan }^{-1}\left(1/3\right)$.