Evaluate the following integrals:

Let $\sin x=t$

We know the Differentiation of $\sin x=\cos x$

$\mathrm{dt}=\mathrm{d}(\sin \mathrm{x})=\cos \mathrm{xd} \mathrm{x}$

So, $\mathrm{dx}=\frac{\mathrm{dt}}{\cos \mathrm{x}}$

substitute all in above equation,

$\int \sin ^{4} x \cos ^{3} x d x=\int t^{4} \cos ^{3} x \frac{d t}{\cos x}$

$=\int t^{4} \cos ^{2} x d t$

$=\int t^{4}\left(1-\sin ^{2} x\right) d t$

$=\int t^{4}\left(1-t^{2}\right) d t$

$=\int\left(t^{4}-t^{6}\right) d t$

We know, basic integration formula, $\int x^{n} d x=\frac{x^{n+1}}{n+1}+c$ for any $c \neq-1$

Hence, $\int\left(t^{4}-t^{6}\right) d t=\frac{t^{5}}{5}-\frac{t^{7}}{7}+c$

Put back $t=\sin x$

$\int \sin ^{4} x \cos ^{3} x d x=\frac{1}{5} \sin ^{5} x-\frac{1}{7} \sin ^{7} x+c$