Question

Let $g\left(x\right)=\int \frac{1+2\cos x}{(\cos x+2{)}^2}dx$ and $g\left(0\right)=0$, the value of $8g\left(\frac{\pi }{2}\right)$ is

Answer 1

$g\left(x\right)=\int \frac{\cos x(\cos x+2)+{\sin }^{2}x}{(\cos x+2{)}^2}dx$
$=\int \underset{II}{\underset{}{\cos x}}.\frac{1}{\underset{I}{\underset{}{\cos x+2}}}dx+\int \frac{{\sin }^{2}x}{(\cos x+2{)}^2}dx$
$=\frac{1}{\cos x+2}\sin x-\int \frac{{\sin }^{2}x}{(\cos x+2{)}^2}dx+\int \frac{{\sin }^{2}x}{(\cos x+2{)}^2}dx$
$\therefore g\left(x\right)=\frac{\sin x}{\cos x+2}+c$
$g\left(0\right)=0\Rightarrow c=0$
$\therefore g\left(x\right)=\frac{\sin x}{\cos x+2}$
$\Rightarrow g\left(\frac{\pi }{2}\right)=\frac{1}{2}$

$\therefore 8g\left(\frac{\pi }{2}\right)=4$