Question

Let $g\left(x\right)=\frac{1+2\cos x}{(\cos x+2{)}^2}dx$ and $g\left(0\right)=0$ then value of $32g\left(\frac{\pi }{2}\right)$ is?

Answer 1

$g\left(x\right)=\int \frac{1+2\cos x}{(\cos x+2{)}^2}dx$

Let $t=\tan \frac{x}{2}$

$dt=\frac{1}{2}{\sec }^2\frac{x}{2}dx=\frac{1}{2}(t^2+1)dx$

$\sin x=\frac{2t}{1+t^2},\cos x=\frac{1-t^2}{1+t^2}$

$2\int \frac{3-t^2}{(t^2+3{)}^2}dt$

$⟹\frac{2t}{t^2+3}+C=\frac{2\tan \frac{x}{2}}{{\tan }^2\frac{x}{2}+3}+C$

At x = 0, we have g(0) = 0,or

$C=0$

At $x=\frac{\pi }{2}$, we have

$\frac{2}{1+3}.32=16$