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

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

Simplify the integrand:

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

Rewrite the sine using the double angle formula $$$\sin\left(2 x\right)=2\sin\left(\frac{2 x}{2}\right)\cos\left(\frac{2 x}{2}\right)$$$:

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

Multiply the numerator and denominator by $$$\sec^2\left(x \right)$$$:

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

Let $$$u=\tan{\left(x \right)}$$$.

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

Therefore,

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

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 $$$u=\tan{\left(x \right)}$$$:

$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \ln{\left(\left|{{\color{red}{\tan{\left(x \right)}}}}\right| \right)}$$

Therefore,

$$\int{\frac{\sec{\left(x \right)}}{\sin{\left(x \right)}} d x} = \ln{\left(\left|{\tan{\left(x \right)}}\right| \right)}$$

Add the constant of integration:

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

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