Integral of $$$\frac{1}{\sin^{2}{\left(\frac{x}{3} \right)}}$$$

Find $$$\int \frac{1}{\sin^{2}{\left(\frac{x}{3} \right)}}\, dx$$$.

Let $$$u=\frac{x}{3}$$$.

Then $$$du=\left(\frac{x}{3}\right)^{\prime }dx = \frac{dx}{3}$$$ (steps can be seen »), and we have that $$$dx = 3 du$$$.

So,

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=3$$$ and $$$f{\left(u \right)} = \frac{1}{\sin^{2}{\left(u \right)}}$$$:

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

Rewrite the integrand in terms of the cosecant:

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

The integral of $$$\csc^{2}{\left(u \right)}$$$ is $$$\int{\csc^{2}{\left(u \right)} d u} = - \cot{\left(u \right)}$$$:

$$3 {\color{red}{\int{\csc^{2}{\left(u \right)} d u}}} = 3 {\color{red}{\left(- \cot{\left(u \right)}\right)}}$$

Recall that $$$u=\frac{x}{3}$$$:

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

Therefore,

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

Add the constant of integration:

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

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