Integral of $$$\frac{1}{\sin{\left(x \right)} \cos^{2}{\left(x \right)}}$$$

Find $$$\int \frac{1}{\sin{\left(x \right)} \cos^{2}{\left(x \right)}}\, dx$$$.

Multiply the numerator and denominator by one sine and write everything else in terms of the cosine, using the formula $$$\sin^2\left(\alpha \right)=-\cos^2\left(\alpha \right)+1$$$ with $$$\alpha=x$$$:

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

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

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

Therefore,

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=-1$$$ and $$$f{\left(u \right)} = \frac{1}{u^{2} \left(1 - u^{2}\right)}$$$:

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

Perform partial fraction decomposition (steps can be seen »):

$$- {\color{red}{\int{\frac{1}{u^{2} \left(1 - u^{2}\right)} d u}}} = - {\color{red}{\int{\left(\frac{1}{2 \left(u + 1\right)} - \frac{1}{2 \left(u - 1\right)} + \frac{1}{u^{2}}\right)d u}}}$$

Integrate term by term:

$$- {\color{red}{\int{\left(\frac{1}{2 \left(u + 1\right)} - \frac{1}{2 \left(u - 1\right)} + \frac{1}{u^{2}}\right)d u}}} = - {\color{red}{\left(\int{\frac{1}{u^{2}} d u} - \int{\frac{1}{2 \left(u - 1\right)} d u} + \int{\frac{1}{2 \left(u + 1\right)} d u}\right)}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-2$$$:

$$\int{\frac{1}{2 \left(u - 1\right)} d u} - \int{\frac{1}{2 \left(u + 1\right)} d u} - {\color{red}{\int{\frac{1}{u^{2}} d u}}}=\int{\frac{1}{2 \left(u - 1\right)} d u} - \int{\frac{1}{2 \left(u + 1\right)} d u} - {\color{red}{\int{u^{-2} d u}}}=\int{\frac{1}{2 \left(u - 1\right)} d u} - \int{\frac{1}{2 \left(u + 1\right)} d u} - {\color{red}{\frac{u^{-2 + 1}}{-2 + 1}}}=\int{\frac{1}{2 \left(u - 1\right)} d u} - \int{\frac{1}{2 \left(u + 1\right)} d u} - {\color{red}{\left(- u^{-1}\right)}}=\int{\frac{1}{2 \left(u - 1\right)} d u} - \int{\frac{1}{2 \left(u + 1\right)} d u} - {\color{red}{\left(- \frac{1}{u}\right)}}$$

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

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

Let $$$v=u + 1$$$.

Then $$$dv=\left(u + 1\right)^{\prime }du = 1 du$$$ (steps can be seen »), and we have that $$$du = dv$$$.

The integral can be rewritten as

$$\int{\frac{1}{2 \left(u - 1\right)} d u} - \frac{{\color{red}{\int{\frac{1}{u + 1} d u}}}}{2} + \frac{1}{u} = \int{\frac{1}{2 \left(u - 1\right)} d u} - \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{2} + \frac{1}{u}$$

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

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

Recall that $$$v=u + 1$$$:

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

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

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

Let $$$v=u - 1$$$.

Then $$$dv=\left(u - 1\right)^{\prime }du = 1 du$$$ (steps can be seen »), and we have that $$$du = dv$$$.

Therefore,

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

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

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

Recall that $$$v=u - 1$$$:

$$- \frac{\ln{\left(\left|{u + 1}\right| \right)}}{2} + \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{2} + \frac{1}{u} = - \frac{\ln{\left(\left|{u + 1}\right| \right)}}{2} + \frac{\ln{\left(\left|{{\color{red}{\left(u - 1\right)}}}\right| \right)}}{2} + \frac{1}{u}$$

Recall that $$$u=\cos{\left(x \right)}$$$:

$$\frac{\ln{\left(\left|{-1 + {\color{red}{u}}}\right| \right)}}{2} - \frac{\ln{\left(\left|{1 + {\color{red}{u}}}\right| \right)}}{2} + {\color{red}{u}}^{-1} = \frac{\ln{\left(\left|{-1 + {\color{red}{\cos{\left(x \right)}}}}\right| \right)}}{2} - \frac{\ln{\left(\left|{1 + {\color{red}{\cos{\left(x \right)}}}}\right| \right)}}{2} + {\color{red}{\cos{\left(x \right)}}}^{-1}$$

Therefore,

$$\int{\frac{1}{\sin{\left(x \right)} \cos^{2}{\left(x \right)}} d x} = \frac{\ln{\left(\left|{\cos{\left(x \right)} - 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{\cos{\left(x \right)} + 1}\right| \right)}}{2} + \frac{1}{\cos{\left(x \right)}}$$

Add the constant of integration:

$$\int{\frac{1}{\sin{\left(x \right)} \cos^{2}{\left(x \right)}} d x} = \frac{\ln{\left(\left|{\cos{\left(x \right)} - 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{\cos{\left(x \right)} + 1}\right| \right)}}{2} + \frac{1}{\cos{\left(x \right)}}+C$$

$$$\int \frac{1}{\sin{\left(x \right)} \cos^{2}{\left(x \right)}}\, dx = \left(\frac{\ln\left(\left|{\cos{\left(x \right)} - 1}\right|\right)}{2} - \frac{\ln\left(\left|{\cos{\left(x \right)} + 1}\right|\right)}{2} + \frac{1}{\cos{\left(x \right)}}\right) + C$$$A