Integral of $$$\sin{\left(x \right)} \cos^{2}{\left(x \right)}$$$

Find $$$\int \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$$$.

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 $$$\sin{\left(x \right)} dx = - du$$$.

The integral can be rewritten as

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

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

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

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=2$$$:

$$- {\color{red}{\int{u^{2} d u}}}=- {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=- {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

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

$$- \frac{{\color{red}{u}}^{3}}{3} = - \frac{{\color{red}{\cos{\left(x \right)}}}^{3}}{3}$$

Therefore,

$$\int{\sin{\left(x \right)} \cos^{2}{\left(x \right)} d x} = - \frac{\cos^{3}{\left(x \right)}}{3}$$

Add the constant of integration:

$$\int{\sin{\left(x \right)} \cos^{2}{\left(x \right)} d x} = - \frac{\cos^{3}{\left(x \right)}}{3}+C$$

$$$\int \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = - \frac{\cos^{3}{\left(x \right)}}{3} + C$$$A