# How do you find the integral of  sin^3[x]dx?

Mar 19, 2016

$\int {\sin}^{3} \left(x\right) \mathrm{dx} = \frac{1}{3} {\cos}^{3} \left(x\right) - \cos \left(x\right) + C$

#### Explanation:

$\int {\sin}^{3} \left(x\right) \mathrm{dx} = \int \sin \left(x\right) \left(1 - {\cos}^{2} \left(x\right)\right) \mathrm{dx}$

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

For the first integral:

$\int \sin \left(x\right) \mathrm{dx} = - \cos \left(x\right) + C$

For the second integral, using substitution:

Let $u = \cos \left(x\right) \implies \mathrm{du} = - \sin \left(x\right) \mathrm{dx}$
Then

$- \int \sin \left(x\right) {\cos}^{2} \left(x\right) \mathrm{dx} = \int {u}^{2} \mathrm{du}$

$= {u}^{3} / 3 + C$

$= \frac{1}{3} {\cos}^{3} \left(x\right) + C$

Putting it all together, we get our final result:

$\int {\sin}^{3} \left(x\right) \mathrm{dx} = \int \sin \left(x\right) \mathrm{dx} - \int \sin \left(x\right) {\cos}^{2} \left(x\right) \mathrm{dx}$

$= - \cos \left(x\right) + \frac{1}{3} {\cos}^{3} \left(x\right) + C$