What is the integral of sin^5(x)?

Mar 30, 2018

$\int {\sin}^{5} x \mathrm{dx} = {\sin}^{2} x - \sin x - \frac{1}{3} {\sin}^{3} x + C$

Explanation:

We want $\int {\sin}^{5} x \mathrm{dx}$

Rewrite the integrand as $\sin x {\left({\sin}^{2} x\right)}^{2}$:

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

Recall the identity ${\sin}^{2} x + {\cos}^{2} x = 1$. The identity also tells us that

${\sin}^{2} x = 1 - {\cos}^{2} x$

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

This can be solved using a simple substitution. Let

$u = \cos x , \mathrm{du} = - \sin x \mathrm{dx} , - \mathrm{du} = \sin x \mathrm{dx}$

Rewrite the integral:

$- \int {\left(1 - {u}^{2}\right)}^{2} = - \int \left(1 - 2 u + {u}^{2}\right)$

Integrate:

$- \int \left(1 - 2 u + {u}^{2}\right) = - \left(u - {u}^{2} + \frac{1}{3} {u}^{3}\right) + C = {u}^{2} - u - \frac{1}{3} {u}^{3} + C$

Rewriting in terms of $x$ yields

$\int {\sin}^{5} x \mathrm{dx} = {\sin}^{2} x - \sin x - \frac{1}{3} {\sin}^{3} x + C$