Integral of $$$18 \pi^{2} \tan{\left(18 x \right)}$$$

Find $$$\int 18 \pi^{2} \tan{\left(18 x \right)}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=18 \pi^{2}$$$ and $$$f{\left(x \right)} = \tan{\left(18 x \right)}$$$:

$${\color{red}{\int{18 \pi^{2} \tan{\left(18 x \right)} d x}}} = {\color{red}{\left(18 \pi^{2} \int{\tan{\left(18 x \right)} d x}\right)}}$$

Let $$$u=18 x$$$.

Then $$$du=\left(18 x\right)^{\prime }dx = 18 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{18}$$$.

The integral becomes

$$18 \pi^{2} {\color{red}{\int{\tan{\left(18 x \right)} d x}}} = 18 \pi^{2} {\color{red}{\int{\frac{\tan{\left(u \right)}}{18} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{18}$$$ and $$$f{\left(u \right)} = \tan{\left(u \right)}$$$:

$$18 \pi^{2} {\color{red}{\int{\frac{\tan{\left(u \right)}}{18} d u}}} = 18 \pi^{2} {\color{red}{\left(\frac{\int{\tan{\left(u \right)} d u}}{18}\right)}}$$

Rewrite the tangent as $$$\tan\left( u \right)=\frac{\sin\left( u \right)}{\cos\left( u \right)}$$$:

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

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

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

Thus,

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

Apply the constant multiple rule $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ with $$$c=-1$$$ and $$$f{\left(v \right)} = \frac{1}{v}$$$:

$$\pi^{2} {\color{red}{\int{\left(- \frac{1}{v}\right)d v}}} = \pi^{2} {\color{red}{\left(- \int{\frac{1}{v} d v}\right)}}$$

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

$$- \pi^{2} {\color{red}{\int{\frac{1}{v} d v}}} = - \pi^{2} {\color{red}{\ln{\left(\left|{v}\right| \right)}}}$$

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

$$- \pi^{2} \ln{\left(\left|{{\color{red}{v}}}\right| \right)} = - \pi^{2} \ln{\left(\left|{{\color{red}{\cos{\left(u \right)}}}}\right| \right)}$$

Recall that $$$u=18 x$$$:

$$- \pi^{2} \ln{\left(\left|{\cos{\left({\color{red}{u}} \right)}}\right| \right)} = - \pi^{2} \ln{\left(\left|{\cos{\left({\color{red}{\left(18 x\right)}} \right)}}\right| \right)}$$

Therefore,

$$\int{18 \pi^{2} \tan{\left(18 x \right)} d x} = - \pi^{2} \ln{\left(\left|{\cos{\left(18 x \right)}}\right| \right)}$$

Add the constant of integration:

$$\int{18 \pi^{2} \tan{\left(18 x \right)} d x} = - \pi^{2} \ln{\left(\left|{\cos{\left(18 x \right)}}\right| \right)}+C$$

$$$\int 18 \pi^{2} \tan{\left(18 x \right)}\, dx = - \pi^{2} \ln\left(\left|{\cos{\left(18 x \right)}}\right|\right) + C$$$A