Evaluate the following integrals:
$\int \sin ^{4} x \cos ^{3} x d x$
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 $t=\sin x$
$\int \sin ^{4} x \cos ^{3} x d x=\frac{1}{5} \sin ^{5} x-\frac{1}{7} \sin ^{7} x+c$
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