Integral of $$$\frac{\sec{\left(x \right)}}{2}$$$

Find $$$\int \frac{\sec{\left(x \right)}}{2}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(x \right)} = \sec{\left(x \right)}$$$:

$${\color{red}{\int{\frac{\sec{\left(x \right)}}{2} d x}}} = {\color{red}{\left(\frac{\int{\sec{\left(x \right)} d x}}{2}\right)}}$$

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

$$\frac{{\color{red}{\int{\sec{\left(x \right)} d x}}}}{2} = \frac{{\color{red}{\int{\frac{1}{\cos{\left(x \right)}} d x}}}}{2}$$

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

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

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

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

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

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

Therefore,

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

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

$$\frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2} = \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$

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

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

Therefore,

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

Add the constant of integration:

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

$$$\int \frac{\sec{\left(x \right)}}{2}\, dx = \frac{\ln\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right|\right)}{2} + C$$$A