Integral of $$$\csc{\left(3 x \right)}$$$

Find $$$\int \csc{\left(3 x \right)}\, dx$$$.

Let $$$u=3 x$$$.

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

Therefore,

$${\color{red}{\int{\csc{\left(3 x \right)} d x}}} = {\color{red}{\int{\frac{\csc{\left(u \right)}}{3} d u}}}$$

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

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

Rewrite the cosecant as $$$\csc\left( u \right)=\frac{1}{\sin\left( u \right)}$$$:

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

Rewrite the sine using the double angle formula $$$\sin\left( u \right)=2\sin\left(\frac{ u }{2}\right)\cos\left(\frac{ u }{2}\right)$$$:

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

Multiply the numerator and denominator by $$$\sec^2\left(\frac{ u }{2} \right)$$$:

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

Let $$$v=\tan{\left(\frac{u}{2} \right)}$$$.

Then $$$dv=\left(\tan{\left(\frac{u}{2} \right)}\right)^{\prime }du = \frac{\sec^{2}{\left(\frac{u}{2} \right)}}{2} du$$$ (steps can be seen »), and we have that $$$\sec^{2}{\left(\frac{u}{2} \right)} du = 2 dv$$$.

Therefore,

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

The integral of $$$\frac{1}{v}$$$ is $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

$$\frac{{\color{red}{\int{\frac{1}{v} d v}}}}{3} = \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{3}$$

Recall that $$$v=\tan{\left(\frac{u}{2} \right)}$$$:

$$\frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{3} = \frac{\ln{\left(\left|{{\color{red}{\tan{\left(\frac{u}{2} \right)}}}}\right| \right)}}{3}$$

Recall that $$$u=3 x$$$:

$$\frac{\ln{\left(\left|{\tan{\left(\frac{{\color{red}{u}}}{2} \right)}}\right| \right)}}{3} = \frac{\ln{\left(\left|{\tan{\left(\frac{{\color{red}{\left(3 x\right)}}}{2} \right)}}\right| \right)}}{3}$$

Therefore,

$$\int{\csc{\left(3 x \right)} d x} = \frac{\ln{\left(\left|{\tan{\left(\frac{3 x}{2} \right)}}\right| \right)}}{3}$$

Add the constant of integration:

$$\int{\csc{\left(3 x \right)} d x} = \frac{\ln{\left(\left|{\tan{\left(\frac{3 x}{2} \right)}}\right| \right)}}{3}+C$$

$$$\int \csc{\left(3 x \right)}\, dx = \frac{\ln\left(\left|{\tan{\left(\frac{3 x}{2} \right)}}\right|\right)}{3} + C$$$A