Integral of $$$\frac{\sec^{2}{\left(\frac{x}{2} \right)}}{2}$$$
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Find $$$\int \frac{\sec^{2}{\left(\frac{x}{2} \right)}}{2}\, dx$$$.
Solution
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 $$$f{\left(x \right)} = \sec^{2}{\left(\frac{x}{2} \right)}$$$:
$${\color{red}{\int{\frac{\sec^{2}{\left(\frac{x}{2} \right)}}{2} d x}}} = {\color{red}{\left(\frac{\int{\sec^{2}{\left(\frac{x}{2} \right)} d x}}{2}\right)}}$$
Let $$$u=\frac{x}{2}$$$.
Then $$$du=\left(\frac{x}{2}\right)^{\prime }dx = \frac{dx}{2}$$$ (steps can be seen »), and we have that $$$dx = 2 du$$$.
Therefore,
$$\frac{{\color{red}{\int{\sec^{2}{\left(\frac{x}{2} \right)} d x}}}}{2} = \frac{{\color{red}{\int{2 \sec^{2}{\left(u \right)} d u}}}}{2}$$
Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=2$$$ and $$$f{\left(u \right)} = \sec^{2}{\left(u \right)}$$$:
$$\frac{{\color{red}{\int{2 \sec^{2}{\left(u \right)} d u}}}}{2} = \frac{{\color{red}{\left(2 \int{\sec^{2}{\left(u \right)} d u}\right)}}}{2}$$
The integral of $$$\sec^{2}{\left(u \right)}$$$ is $$$\int{\sec^{2}{\left(u \right)} d u} = \tan{\left(u \right)}$$$:
$${\color{red}{\int{\sec^{2}{\left(u \right)} d u}}} = {\color{red}{\tan{\left(u \right)}}}$$
Recall that $$$u=\frac{x}{2}$$$:
$$\tan{\left({\color{red}{u}} \right)} = \tan{\left({\color{red}{\left(\frac{x}{2}\right)}} \right)}$$
Therefore,
$$\int{\frac{\sec^{2}{\left(\frac{x}{2} \right)}}{2} d x} = \tan{\left(\frac{x}{2} \right)}$$
Add the constant of integration:
$$\int{\frac{\sec^{2}{\left(\frac{x}{2} \right)}}{2} d x} = \tan{\left(\frac{x}{2} \right)}+C$$
Answer
$$$\int \frac{\sec^{2}{\left(\frac{x}{2} \right)}}{2}\, dx = \tan{\left(\frac{x}{2} \right)} + C$$$A