Integral of $$$\sin^{9}{\left(\theta \right)} \cos{\left(\theta \right)}$$$

Find $$$\int \sin^{9}{\left(\theta \right)} \cos{\left(\theta \right)}\, d\theta$$$.

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

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

The integral can be rewritten as

$${\color{red}{\int{\sin^{9}{\left(\theta \right)} \cos{\left(\theta \right)} d \theta}}} = {\color{red}{\int{u^{9} d u}}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=9$$$:

$${\color{red}{\int{u^{9} d u}}}={\color{red}{\frac{u^{1 + 9}}{1 + 9}}}={\color{red}{\left(\frac{u^{10}}{10}\right)}}$$

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

$$\frac{{\color{red}{u}}^{10}}{10} = \frac{{\color{red}{\sin{\left(\theta \right)}}}^{10}}{10}$$

Therefore,

$$\int{\sin^{9}{\left(\theta \right)} \cos{\left(\theta \right)} d \theta} = \frac{\sin^{10}{\left(\theta \right)}}{10}$$

Add the constant of integration:

$$\int{\sin^{9}{\left(\theta \right)} \cos{\left(\theta \right)} d \theta} = \frac{\sin^{10}{\left(\theta \right)}}{10}+C$$

$$$\int \sin^{9}{\left(\theta \right)} \cos{\left(\theta \right)}\, d\theta = \frac{\sin^{10}{\left(\theta \right)}}{10} + C$$$A