# Evaluate the following integral:

Question:

Evaluate the following integral:

$\int \frac{1}{\sin x(3+2 \cos x)} d x$

Solution:

Let, $\mathrm{I}=\int \frac{1}{\sin \mathrm{x}(3+2 \cos \mathrm{x})} \mathrm{dx}$

Multiplying and dividing by $\sin x$

$\therefore \mathrm{I}=\int \frac{\sin \mathrm{x}}{\sin ^{2} \mathrm{x}(3+2 \cos \mathrm{x})} \mathrm{dx}$

$\therefore \mathrm{I}=\int \frac{\sin \mathrm{x}}{\left(1-\cos ^{2} \mathrm{x}\right)(3+2 \cos \mathrm{x})} \mathrm{dx}$

Let $\cos x=t,-\sin x d x=d t$

So, $I=\int \frac{d t}{\left(t^{2}-1\right)(3+2 t)}$

Now, let $\frac{1}{\left(t^{2}-1\right)(3+2 t)}=\frac{A}{t-1}+\frac{B}{t+1}+\frac{C}{3+2 t}$

$\Rightarrow 1=A(t+1)(3+2 t)+B(t-1)(3+2 t)+C\left(t^{2}-1\right)$

For, $\mathrm{t}=1, \mathrm{~A}=\frac{1}{10}$

For, $t=-1, B=-\frac{1}{2}$

For, $t=-\frac{3}{2}, C=\frac{4}{5}$

$\therefore \mathrm{I}=\frac{1}{10} \int \frac{\mathrm{dt}}{\mathrm{t}-1}-\frac{1}{2} \int \frac{\mathrm{dt}}{\mathrm{t}+1}+\frac{4}{5} \int \frac{\mathrm{dt}}{3+2 \mathrm{t}}$

$\Rightarrow I=\frac{1}{10} \log |t-1|-\frac{1}{2} \log |t+1|+\frac{2}{5} \log |3+2 t|+c$