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

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

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

Therefore,

$${\color{red}{\int{\frac{\cos{\left(x \right)}}{\sin^{3}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{1}{u^{3}} d u}}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-3$$$:

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

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

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

Therefore,

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

Add the constant of integration:

$$\int{\frac{\cos{\left(x \right)}}{\sin^{3}{\left(x \right)}} d x} = - \frac{1}{2 \sin^{2}{\left(x \right)}}+C$$

$$$\int \frac{\cos{\left(x \right)}}{\sin^{3}{\left(x \right)}}\, dx = - \frac{1}{2 \sin^{2}{\left(x \right)}} + C$$$A