Integral of $$$\sin^{5}{\left(x \right)} \cos{\left(x \right)}$$$

Find $$$\int \sin^{5}{\left(x \right)} \cos{\left(x \right)}\, dx$$$.

Let $$$u=\sin{\left(x \right)}$$$.

Then $$$du=\left(\sin{\left(x \right)}\right)^{\prime }dx = \cos{\left(x \right)} dx$$$ (steps can be seen »), and we have that $$$\cos{\left(x \right)} dx = du$$$.

The integral can be rewritten as

$${\color{red}{\int{\sin^{5}{\left(x \right)} \cos{\left(x \right)} d x}}} = {\color{red}{\int{u^{5} d u}}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=5$$$:

$${\color{red}{\int{u^{5} d u}}}={\color{red}{\frac{u^{1 + 5}}{1 + 5}}}={\color{red}{\left(\frac{u^{6}}{6}\right)}}$$

Recall that $$$u=\sin{\left(x \right)}$$$:

$$\frac{{\color{red}{u}}^{6}}{6} = \frac{{\color{red}{\sin{\left(x \right)}}}^{6}}{6}$$

Therefore,

$$\int{\sin^{5}{\left(x \right)} \cos{\left(x \right)} d x} = \frac{\sin^{6}{\left(x \right)}}{6}$$

Add the constant of integration:

$$\int{\sin^{5}{\left(x \right)} \cos{\left(x \right)} d x} = \frac{\sin^{6}{\left(x \right)}}{6}+C$$

$$$\int \sin^{5}{\left(x \right)} \cos{\left(x \right)}\, dx = \frac{\sin^{6}{\left(x \right)}}{6} + C$$$A