Evaluate the following integral:

Solution:

Let

$\sin x=t$

$\cos x d x=d t$

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

$=\int \frac{d t}{(1-t)^{3}(2+t)}$

$\frac{1}{(1-t)^{3}(2+t)}=\frac{A}{1-t}+\frac{B}{(1-t)^{2}}+\frac{C}{(1-t)^{3}}+\frac{D}{2+t}$

$1=A(1-t)^{2}(2+t)+B(1-t)(2+t)+C(2+t)+D(1-t)^{3}$

Put $\mathrm{t}=1$

$1=3 C$

$C=\frac{1}{3}$

Put $t=-2$

$1=27 \mathrm{D}$

$D=\frac{1}{27}$

$A=-\frac{1}{27} B=\frac{1}{9}$

$\int \frac{d t}{(1-t)^{3}(2+t)}$

$\quad=-\frac{1}{27} \int \frac{1}{1-t} d t+\frac{1}{9} \int \frac{d t}{(1-t)^{2}}+\frac{1}{3} \int \frac{d t}{(1-t)^{3}}+\frac{1}{27} \int \frac{d t}{2+t}$

$=-\frac{1}{27} \log |1-t|+\frac{1}{9(1-t)}+\frac{1}{6(1-t)^{2}}+\frac{1}{27} \log |2+t|+C$

Put $t=\sin x$

$=-\frac{1}{27} \log |1-\sin x|+\frac{1}{9(1-\sin x)}+\frac{1}{6(1-\sin x)^{2}}+\frac{1}{27} \log |2+\sin x|$

$+C$