Integral of $$$6 \cos^{2}{\left(x \right)}$$$

Find $$$\int 6 \cos^{2}{\left(x \right)}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=6$$$ and $$$f{\left(x \right)} = \cos^{2}{\left(x \right)}$$$:

$${\color{red}{\int{6 \cos^{2}{\left(x \right)} d x}}} = {\color{red}{\left(6 \int{\cos^{2}{\left(x \right)} d x}\right)}}$$

Apply the power reducing formula $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$ with $$$\alpha=x$$$:

$$6 {\color{red}{\int{\cos^{2}{\left(x \right)} d x}}} = 6 {\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)d x}}}$$

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 $$$f{\left(x \right)} = \cos{\left(2 x \right)} + 1$$$:

$$6 {\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)d x}}} = 6 {\color{red}{\left(\frac{\int{\left(\cos{\left(2 x \right)} + 1\right)d x}}{2}\right)}}$$

Integrate term by term:

$$3 {\color{red}{\int{\left(\cos{\left(2 x \right)} + 1\right)d x}}} = 3 {\color{red}{\left(\int{1 d x} + \int{\cos{\left(2 x \right)} d x}\right)}}$$

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=1$$$:

$$3 \int{\cos{\left(2 x \right)} d x} + 3 {\color{red}{\int{1 d x}}} = 3 \int{\cos{\left(2 x \right)} d x} + 3 {\color{red}{x}}$$

Let $$$u=2 x$$$.

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

Therefore,

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

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 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

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

The integral of the cosine is $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:

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

Recall that $$$u=2 x$$$:

$$3 x + \frac{3 \sin{\left({\color{red}{u}} \right)}}{2} = 3 x + \frac{3 \sin{\left({\color{red}{\left(2 x\right)}} \right)}}{2}$$

Therefore,

$$\int{6 \cos^{2}{\left(x \right)} d x} = 3 x + \frac{3 \sin{\left(2 x \right)}}{2}$$

Add the constant of integration:

$$\int{6 \cos^{2}{\left(x \right)} d x} = 3 x + \frac{3 \sin{\left(2 x \right)}}{2}+C$$

$$$\int 6 \cos^{2}{\left(x \right)}\, dx = \left(3 x + \frac{3 \sin{\left(2 x \right)}}{2}\right) + C$$$A