Integral of $$$\frac{1}{\cos{\left(3 x \right)}}$$$

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

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

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

Thus,

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

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

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

Rewrite the cosine in terms of the sine using the formula $$$\cos\left( u \right)=\sin\left( u + \frac{\pi}{2}\right)$$$ and then rewrite the sine using the double angle formula $$$\sin\left( u \right)=2\sin\left(\frac{ u }{2}\right)\cos\left(\frac{ u }{2}\right)$$$:

$$\frac{{\color{red}{\int{\frac{1}{\cos{\left(u \right)}} d u}}}}{3} = \frac{{\color{red}{\int{\frac{1}{2 \sin{\left(\frac{u}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{3}$$

Multiply the numerator and denominator by $$$\sec^2\left(\frac{ u }{2} + \frac{\pi}{4} \right)$$$:

$$\frac{{\color{red}{\int{\frac{1}{2 \sin{\left(\frac{u}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{3} = \frac{{\color{red}{\int{\frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{3}$$

Let $$$v=\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}$$$.

Then $$$dv=\left(\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}\right)^{\prime }du = \frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}{2} du$$$ (steps can be seen »), and we have that $$$\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)} du = 2 dv$$$.

Therefore,

$$\frac{{\color{red}{\int{\frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{3} = \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{3}$$

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 $$$v=\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}$$$:

$$\frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{3} = \frac{\ln{\left(\left|{{\color{red}{\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}}}\right| \right)}}{3}$$

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

$$\frac{\ln{\left(\left|{\tan{\left(\frac{\pi}{4} + \frac{{\color{red}{u}}}{2} \right)}}\right| \right)}}{3} = \frac{\ln{\left(\left|{\tan{\left(\frac{\pi}{4} + \frac{{\color{red}{\left(3 x\right)}}}{2} \right)}}\right| \right)}}{3}$$

Therefore,

$$\int{\frac{1}{\cos{\left(3 x \right)}} d x} = \frac{\ln{\left(\left|{\tan{\left(\frac{3 x}{2} + \frac{\pi}{4} \right)}}\right| \right)}}{3}$$

Simplify:

$$\int{\frac{1}{\cos{\left(3 x \right)}} d x} = \frac{\ln{\left(\left|{\tan{\left(\frac{6 x + \pi}{4} \right)}}\right| \right)}}{3}$$

Add the constant of integration:

$$\int{\frac{1}{\cos{\left(3 x \right)}} d x} = \frac{\ln{\left(\left|{\tan{\left(\frac{6 x + \pi}{4} \right)}}\right| \right)}}{3}+C$$

$$$\int \frac{1}{\cos{\left(3 x \right)}}\, dx = \frac{\ln\left(\left|{\tan{\left(\frac{6 x + \pi}{4} \right)}}\right|\right)}{3} + C$$$A