Encontrar $\displaystyle\int \sin^3 x\ dx$.

$I\ =\ \displaystyle\int \sin^3 x\ dx\ =\ \displaystyle\int (\sin x)(1 - \cos^2 x) x\ dx\ =$

$=\ -(\cos x)(1 - \cos^2 x) + 2\displaystyle\int (\cos^2 x)(\sin x)\ dx\ =$

$=\ -\cos x + \cos^3 x + 2\displaystyle\int (1 - \sin^2 x)(\sin x)\ dx\ =$

$=\ -\cos x + \cos^3 x + 2\displaystyle\int \sin x\ dx - 2\underset{I}{\underbrace{\displaystyle\int\sin^3 x\ dx}}$

$\fbox{$\displaystyle\int \sin^3 x\ dx\ =\ \dfrac{\cos^3 x}{3} - \cos x + c$}$