How do I evaluate the indefinite integral intsin^6(x)*cos^3(x)dx ?

Sep 21, 2014

By substitution,

$\int {\sin}^{6} x {\cos}^{3} x \mathrm{dx} = \frac{{\sin}^{7} x}{7} - \frac{{\sin}^{9} x}{9} + C$

Let us look at some details.

$\int {\sin}^{6} x {\cos}^{3} x \mathrm{dx}$

by pulling out $\cos x$,

$= \int {\sin}^{6} x {\cos}^{2} x \cdot \cos x \mathrm{dx}$

by the trig identity ${\cos}^{2} x = 1 - {\sin}^{2} x$,

$= \int {\sin}^{6} x \left(1 - {\sin}^{2} x\right) \cdot \cos x \mathrm{dx}$

by the substitution $u = \sin x$. $R i g h t a r r o w \mathrm{du} = \cos x \mathrm{dx}$,

$= \int {u}^{6} \left(1 - {u}^{2}\right) \mathrm{du} = \int {u}^{6} - {u}^{8} \mathrm{du}$

by Power Rule,

$= {u}^{7} / 7 - {u}^{9} / 9 + C$

by putting $u = \sin x$ back in,

$= \frac{{\sin}^{7} x}{7} - \frac{{\sin}^{9} x}{9} + C$