Integral of $$$\sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)}$$$

Find $$$\int \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$$$.

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

$${\color{red}{\int{\sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)} d x}}} = {\color{red}{\int{\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{4}{\left(x \right)} d x}}}$$

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

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

Therefore,

$${\color{red}{\int{\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{4}{\left(x \right)} d x}}} = {\color{red}{\int{\left(- u^{4} \left(1 - u^{2}\right)\right)d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=-1$$$ and $$$f{\left(u \right)} = u^{4} \left(1 - u^{2}\right)$$$:

$${\color{red}{\int{\left(- u^{4} \left(1 - u^{2}\right)\right)d u}}} = {\color{red}{\left(- \int{u^{4} \left(1 - u^{2}\right) d u}\right)}}$$

Expand the expression:

$$- {\color{red}{\int{u^{4} \left(1 - u^{2}\right) d u}}} = - {\color{red}{\int{\left(- u^{6} + u^{4}\right)d u}}}$$

Integrate term by term:

$$- {\color{red}{\int{\left(- u^{6} + u^{4}\right)d u}}} = - {\color{red}{\left(\int{u^{4} d u} - \int{u^{6} d u}\right)}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=4$$$:

$$\int{u^{6} d u} - {\color{red}{\int{u^{4} d u}}}=\int{u^{6} d u} - {\color{red}{\frac{u^{1 + 4}}{1 + 4}}}=\int{u^{6} d u} - {\color{red}{\left(\frac{u^{5}}{5}\right)}}$$

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

$$- \frac{u^{5}}{5} + {\color{red}{\int{u^{6} d u}}}=- \frac{u^{5}}{5} + {\color{red}{\frac{u^{1 + 6}}{1 + 6}}}=- \frac{u^{5}}{5} + {\color{red}{\left(\frac{u^{7}}{7}\right)}}$$

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

$$- \frac{{\color{red}{u}}^{5}}{5} + \frac{{\color{red}{u}}^{7}}{7} = - \frac{{\color{red}{\cos{\left(x \right)}}}^{5}}{5} + \frac{{\color{red}{\cos{\left(x \right)}}}^{7}}{7}$$

Therefore,

$$\int{\sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)} d x} = \frac{\cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}$$

Add the constant of integration:

$$\int{\sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)} d x} = \frac{\cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}+C$$

$$$\int \sin^{3}{\left(x \right)} \cos^{4}{\left(x \right)}\, dx = \left(\frac{\cos^{7}{\left(x \right)}}{7} - \frac{\cos^{5}{\left(x \right)}}{5}\right) + C$$$A