Question

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

Answer 1

$\sin \left(x\right)$	t
0	0
$\frac{\pi }{2}$	1

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

Put $\sin x=t$

Differentiating we get,

$\cos xdx=dt$

Refer diagram.

$I={\int }_0^1\frac{1}{(1+t)(2+t)}dt$

$\frac{1}{(1+t)(2+t)}=\frac{A}{1+t}+\frac{B}{2+t}$

$1=A(2+t)+8(1+t)$

Putting $t=-1$ in above equation

$A=1$

Putting $t=-2$ in above equation

$B=-1$

$\therefore \frac{1}{(1+t)(2+t)}=\frac{1}{1+t}-\frac{1}{2+t}$

$\therefore I={\int }_0^1[\frac{1}{1+t}-\frac{1}{2+t}]dt$

$=[\log |1+t|{]}_0^1-[log|2+t|{]}_0^1$

$=[log2-log1]=[log3-log2]$

$=log2-log\left(\frac{3}{2}\right)$

$[\because {\int }_0^{\pi /2}\frac{\cos x}{(1+\sin x)(2+\sin x)}dx=log\left(\frac{4}{3}\right)]$.