Evaluate the following integral:

Solution:

The denominator is factorized, so let separate the fraction through partial fraction, hence let

$\frac{\sin 2 x}{(1+\sin x)(2+\sin x)}=\frac{A}{(1+\sin x)}+\frac{B}{2+\sin x} \ldots \ldots$ (i)

$\Rightarrow \frac{\sin 2 x}{(1+\sin x)(2+\sin x)}=\frac{A(2+\sin x)+B(1+\sin x)}{(1+\sin x)(2+\sin x)}$

$\Rightarrow \sin 2 x=A(2+\sin x)+B(1+\sin x)=2 A+A \sin x+B+B \sin x$

$\Rightarrow 2 \sin x \cos x=\sin x(A+B)+(2 A+B) \ldots \ldots$ (ii)

We need to solve for $A$ and $B$.

We will equate similar terms, we get.

$2 A+B=0 \Rightarrow B=-2 A$

And $A+B=2 \cos x$

Substituting the value of $B$, we get

$A-2 A=2 \cos x \Rightarrow A=-2 \cos x$

Hence $B=-2 A=-2(-2 \cos x)$

$\Rightarrow B=4 \cos x$

We put the values of $A$ and $B$ values back into our partial fractions in equation (i) and replace this as the integrand. We get

$\int\left[\frac{\sin 2 x}{(1+\sin x)(2+\sin x)}\right] d x$

$\Rightarrow \int\left[\frac{A}{(1+\sin x)}+\frac{B}{2+\sin x}\right] d x$

$\Rightarrow \int\left[\frac{-2 \cos x}{(1+\sin x)}+\frac{4 \cos x}{2+\sin x}\right] d x$

Split up the integral,

$\Rightarrow-\int \frac{2 \cos x}{(1+\sin x)} d x+\int \frac{4 \cos x}{2+\sin x} d x$

Let substitute

$u=\sin x \Rightarrow d u=\cos x d x$

so the above equation becomes,

$\Rightarrow-2 \int \frac{1}{(1+u)} d u+4 \int \frac{1}{2+u} d u$

Now substitute

$v=1+u \Rightarrow d v=d u$

$z=2+u \Rightarrow d z=d u$

So above equation becomes,

$\Rightarrow-2 \int \frac{1}{(v)} d v+4 \int \frac{1}{z} d z$

On integrating we get

$\Rightarrow-2 \log |v|+4 \log |z|+C$

Substituting back, we get

$\Rightarrow 4 \log |2+u|-2 \log |1+u|+C$

$\Rightarrow 4 \log |2+\sin x|-2 \log |1+\sin x|+C$

Applying logarithm rule, we get

$\Rightarrow \log \left|(2+\sin x)^{4}\right|-\log \left|(1+\sin x)^{2}\right|+C$

$\Rightarrow \log \left|\frac{(2+\sin x)^{4}}{(1+\sin x)^{2}}\right|+C$

Note: the absolute value signs account for the domain of the natural log function $(x>0)$.

Hence,

$\int \frac{\sin 2 x}{(1+\sin x)(2+\sin x)} d x=\log \left|\frac{(2+\sin x)^{4}}{(1+\sin x)^{2}}\right|+C$