Integral of $$$\sin^{5}{\left(x \right)} \cos^{3}{\left(x \right)}$$$
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Find $$$\int \sin^{5}{\left(x \right)} \cos^{3}{\left(x \right)}\, dx$$$.
Solution
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 $$$\alpha=x$$$:
$${\color{red}{\int{\sin^{5}{\left(x \right)} \cos^{3}{\left(x \right)} d x}}} = {\color{red}{\int{\left(1 - \sin^{2}{\left(x \right)}\right) \sin^{5}{\left(x \right)} \cos{\left(x \right)} d x}}}$$
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 can be rewritten as
$${\color{red}{\int{\left(1 - \sin^{2}{\left(x \right)}\right) \sin^{5}{\left(x \right)} \cos{\left(x \right)} d x}}} = {\color{red}{\int{u^{5} \left(1 - u^{2}\right) d u}}}$$
Expand the expression:
$${\color{red}{\int{u^{5} \left(1 - u^{2}\right) d u}}} = {\color{red}{\int{\left(- u^{7} + u^{5}\right)d u}}}$$
Integrate term by term:
$${\color{red}{\int{\left(- u^{7} + u^{5}\right)d u}}} = {\color{red}{\left(\int{u^{5} d u} - \int{u^{7} d u}\right)}}$$
Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=5$$$:
$$- \int{u^{7} d u} + {\color{red}{\int{u^{5} d u}}}=- \int{u^{7} d u} + {\color{red}{\frac{u^{1 + 5}}{1 + 5}}}=- \int{u^{7} d u} + {\color{red}{\left(\frac{u^{6}}{6}\right)}}$$
Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=7$$$:
$$\frac{u^{6}}{6} - {\color{red}{\int{u^{7} d u}}}=\frac{u^{6}}{6} - {\color{red}{\frac{u^{1 + 7}}{1 + 7}}}=\frac{u^{6}}{6} - {\color{red}{\left(\frac{u^{8}}{8}\right)}}$$
Recall that $$$u=\sin{\left(x \right)}$$$:
$$\frac{{\color{red}{u}}^{6}}{6} - \frac{{\color{red}{u}}^{8}}{8} = \frac{{\color{red}{\sin{\left(x \right)}}}^{6}}{6} - \frac{{\color{red}{\sin{\left(x \right)}}}^{8}}{8}$$
Therefore,
$$\int{\sin^{5}{\left(x \right)} \cos^{3}{\left(x \right)} d x} = - \frac{\sin^{8}{\left(x \right)}}{8} + \frac{\sin^{6}{\left(x \right)}}{6}$$
Add the constant of integration:
$$\int{\sin^{5}{\left(x \right)} \cos^{3}{\left(x \right)} d x} = - \frac{\sin^{8}{\left(x \right)}}{8} + \frac{\sin^{6}{\left(x \right)}}{6}+C$$
Answer
$$$\int \sin^{5}{\left(x \right)} \cos^{3}{\left(x \right)}\, dx = \left(- \frac{\sin^{8}{\left(x \right)}}{8} + \frac{\sin^{6}{\left(x \right)}}{6}\right) + C$$$A