Integral of $$$\sin^{6}{\left(x \right)} \cos^{3}{\left(x \right)}$$$

Find $$$\int \sin^{6}{\left(x \right)} \cos^{3}{\left(x \right)}\, dx$$$.

Strip out one cosine and write everything else in terms of the sine, using the formula $$$\cos^2\left(\alpha \right)=-\sin^2\left(\alpha \right)+1$$$ with $$$\alpha=x$$$:

$${\color{red}{\int{\sin^{6}{\left(x \right)} \cos^{3}{\left(x \right)} d x}}} = {\color{red}{\int{\left(1 - \sin^{2}{\left(x \right)}\right) \sin^{6}{\left(x \right)} \cos{\left(x \right)} d x}}}$$

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 becomes

$${\color{red}{\int{\left(1 - \sin^{2}{\left(x \right)}\right) \sin^{6}{\left(x \right)} \cos{\left(x \right)} d x}}} = {\color{red}{\int{u^{6} \left(1 - u^{2}\right) d u}}}$$

Expand the expression:

$${\color{red}{\int{u^{6} \left(1 - u^{2}\right) d u}}} = {\color{red}{\int{\left(- u^{8} + u^{6}\right)d u}}}$$

Integrate term by term:

$${\color{red}{\int{\left(- u^{8} + u^{6}\right)d u}}} = {\color{red}{\left(\int{u^{6} d u} - \int{u^{8} d u}\right)}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=6$$$:

$$- \int{u^{8} d u} + {\color{red}{\int{u^{6} d u}}}=- \int{u^{8} d u} + {\color{red}{\frac{u^{1 + 6}}{1 + 6}}}=- \int{u^{8} d u} + {\color{red}{\left(\frac{u^{7}}{7}\right)}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=8$$$:

$$\frac{u^{7}}{7} - {\color{red}{\int{u^{8} d u}}}=\frac{u^{7}}{7} - {\color{red}{\frac{u^{1 + 8}}{1 + 8}}}=\frac{u^{7}}{7} - {\color{red}{\left(\frac{u^{9}}{9}\right)}}$$

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

$$\frac{{\color{red}{u}}^{7}}{7} - \frac{{\color{red}{u}}^{9}}{9} = \frac{{\color{red}{\sin{\left(x \right)}}}^{7}}{7} - \frac{{\color{red}{\sin{\left(x \right)}}}^{9}}{9}$$

Therefore,

$$\int{\sin^{6}{\left(x \right)} \cos^{3}{\left(x \right)} d x} = - \frac{\sin^{9}{\left(x \right)}}{9} + \frac{\sin^{7}{\left(x \right)}}{7}$$

Add the constant of integration:

$$\int{\sin^{6}{\left(x \right)} \cos^{3}{\left(x \right)} d x} = - \frac{\sin^{9}{\left(x \right)}}{9} + \frac{\sin^{7}{\left(x \right)}}{7}+C$$

$$$\int \sin^{6}{\left(x \right)} \cos^{3}{\left(x \right)}\, dx = \left(- \frac{\sin^{9}{\left(x \right)}}{9} + \frac{\sin^{7}{\left(x \right)}}{7}\right) + C$$$A