Integral of $$$- \frac{3 \sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} + 2$$$

Find $$$\int \left(- \frac{3 \sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} + 2\right)\, dx$$$.

Integrate term by term:

$${\color{red}{\int{\left(- \frac{3 \sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} + 2\right)d x}}} = {\color{red}{\left(\int{2 d x} - \int{\frac{3 \sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} d x}\right)}}$$

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=2$$$:

$$- \int{\frac{3 \sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} d x} + {\color{red}{\int{2 d x}}} = - \int{\frac{3 \sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} d x} + {\color{red}{\left(2 x\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=3$$$ and $$$f{\left(x \right)} = \frac{\sin{\left(x \right)}}{\cos^{2}{\left(x \right)}}$$$:

$$2 x - {\color{red}{\int{\frac{3 \sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} d x}}} = 2 x - {\color{red}{\left(3 \int{\frac{\sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} d x}\right)}}$$

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 $$$\sin{\left(x \right)} dx = - du$$$.

So,

$$2 x - 3 {\color{red}{\int{\frac{\sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} d x}}} = 2 x - 3 {\color{red}{\int{\left(- \frac{1}{u^{2}}\right)d u}}}$$

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

$$2 x - 3 {\color{red}{\int{\left(- \frac{1}{u^{2}}\right)d u}}} = 2 x - 3 {\color{red}{\left(- \int{\frac{1}{u^{2}} d u}\right)}}$$

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

$$2 x + 3 {\color{red}{\int{\frac{1}{u^{2}} d u}}}=2 x + 3 {\color{red}{\int{u^{-2} d u}}}=2 x + 3 {\color{red}{\frac{u^{-2 + 1}}{-2 + 1}}}=2 x + 3 {\color{red}{\left(- u^{-1}\right)}}=2 x + 3 {\color{red}{\left(- \frac{1}{u}\right)}}$$

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

$$2 x - 3 {\color{red}{u}}^{-1} = 2 x - 3 {\color{red}{\cos{\left(x \right)}}}^{-1}$$

Therefore,

$$\int{\left(- \frac{3 \sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} + 2\right)d x} = 2 x - \frac{3}{\cos{\left(x \right)}}$$

Add the constant of integration:

$$\int{\left(- \frac{3 \sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} + 2\right)d x} = 2 x - \frac{3}{\cos{\left(x \right)}}+C$$

$$$\int \left(- \frac{3 \sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} + 2\right)\, dx = \left(2 x - \frac{3}{\cos{\left(x \right)}}\right) + C$$$A