Solution:

As power of $\sin$ is odd, put $\cos x=t$
Therefore $\mathrm{dt}=-\sin \mathrm{x} \mathrm{d} \mathrm{x}$
On substitute these in above equation, we get
$\begin{array}{l}
\int \sin ^{3} x \cos ^{6} x d x=\int \sin x \sin ^{2} x t^{6} d x \\
=\int\left(1-\cos ^{2} x\right) t^{6} \sin x d x
\end{array}$
$=\int-\left(t^{6}-t^{8}\right) d t$
On integrating we obtain
$=-\frac{\mathrm{t}^{7}}{7}+\frac{\mathrm{t}^{9}}{9}+\mathrm{c}$
Putting the value of $t$ we obtain
$=\frac{-1}{7} \cos ^{7} x+\frac{1}{9} \cos ^{9} x+c$