Integral of $$$\sec{\left(\theta \right)}$$$

Rewrite the secant as $$$\sec\left(\theta\right)=\frac{1}{\cos\left(\theta\right)}$$$:

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

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

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

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

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

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

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

The integral can be rewritten as

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

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

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

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

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

Therefore,

$$\int{\sec{\left(\theta \right)} d \theta} = \ln{\left(\left|{\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}}\right| \right)}$$

Add the constant of integration:

$$\int{\sec{\left(\theta \right)} d \theta} = \ln{\left(\left|{\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}}\right| \right)}+C$$

Answer: $$$\int{\sec{\left(\theta \right)} d \theta}=\ln{\left(\left|{\tan{\left(\frac{\theta}{2} + \frac{\pi}{4} \right)}}\right| \right)}+C$$$