Integral of $$$\cos{\left(2 x \right)} \cos{\left(3 x - 5 \right)}$$$

Find $$$\int \cos{\left(2 x \right)} \cos{\left(3 x - 5 \right)}\, dx$$$.

Rewrite the integrand using the formula $$$\cos\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \cos\left(\alpha-\beta \right)+\frac{1}{2} \cos\left(\alpha+\beta \right)$$$ with $$$\alpha=2 x$$$ and $$$\beta=3 x - 5$$$:

$${\color{red}{\int{\cos{\left(2 x \right)} \cos{\left(3 x - 5 \right)} d x}}} = {\color{red}{\int{\left(\frac{\cos{\left(x - 5 \right)}}{2} + \frac{\cos{\left(5 x - 5 \right)}}{2}\right)d x}}}$$

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 $$$f{\left(x \right)} = \cos{\left(x - 5 \right)} + \cos{\left(5 x - 5 \right)}$$$:

$${\color{red}{\int{\left(\frac{\cos{\left(x - 5 \right)}}{2} + \frac{\cos{\left(5 x - 5 \right)}}{2}\right)d x}}} = {\color{red}{\left(\frac{\int{\left(\cos{\left(x - 5 \right)} + \cos{\left(5 x - 5 \right)}\right)d x}}{2}\right)}}$$

Integrate term by term:

$$\frac{{\color{red}{\int{\left(\cos{\left(x - 5 \right)} + \cos{\left(5 x - 5 \right)}\right)d x}}}}{2} = \frac{{\color{red}{\left(\int{\cos{\left(x - 5 \right)} d x} + \int{\cos{\left(5 x - 5 \right)} d x}\right)}}}{2}$$

Let $$$u=x - 5$$$.

Then $$$du=\left(x - 5\right)^{\prime }dx = 1 dx$$$ (steps can be seen »), and we have that $$$dx = du$$$.

So,

$$\frac{\int{\cos{\left(5 x - 5 \right)} d x}}{2} + \frac{{\color{red}{\int{\cos{\left(x - 5 \right)} d x}}}}{2} = \frac{\int{\cos{\left(5 x - 5 \right)} d x}}{2} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{2}$$

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

$$\frac{\int{\cos{\left(5 x - 5 \right)} d x}}{2} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{2} = \frac{\int{\cos{\left(5 x - 5 \right)} d x}}{2} + \frac{{\color{red}{\sin{\left(u \right)}}}}{2}$$

Recall that $$$u=x - 5$$$:

$$\frac{\int{\cos{\left(5 x - 5 \right)} d x}}{2} + \frac{\sin{\left({\color{red}{u}} \right)}}{2} = \frac{\int{\cos{\left(5 x - 5 \right)} d x}}{2} + \frac{\sin{\left({\color{red}{\left(x - 5\right)}} \right)}}{2}$$

Let $$$u=5 x - 5$$$.

Then $$$du=\left(5 x - 5\right)^{\prime }dx = 5 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{5}$$$.

The integral becomes

$$\frac{\sin{\left(x - 5 \right)}}{2} + \frac{{\color{red}{\int{\cos{\left(5 x - 5 \right)} d x}}}}{2} = \frac{\sin{\left(x - 5 \right)}}{2} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{5} d u}}}}{2}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{5}$$$ and $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$$\frac{\sin{\left(x - 5 \right)}}{2} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{5} d u}}}}{2} = \frac{\sin{\left(x - 5 \right)}}{2} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{5}\right)}}}{2}$$

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

$$\frac{\sin{\left(x - 5 \right)}}{2} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{10} = \frac{\sin{\left(x - 5 \right)}}{2} + \frac{{\color{red}{\sin{\left(u \right)}}}}{10}$$

Recall that $$$u=5 x - 5$$$:

$$\frac{\sin{\left(x - 5 \right)}}{2} + \frac{\sin{\left({\color{red}{u}} \right)}}{10} = \frac{\sin{\left(x - 5 \right)}}{2} + \frac{\sin{\left({\color{red}{\left(5 x - 5\right)}} \right)}}{10}$$

Therefore,

$$\int{\cos{\left(2 x \right)} \cos{\left(3 x - 5 \right)} d x} = \frac{\sin{\left(x - 5 \right)}}{2} + \frac{\sin{\left(5 x - 5 \right)}}{10}$$

Add the constant of integration:

$$\int{\cos{\left(2 x \right)} \cos{\left(3 x - 5 \right)} d x} = \frac{\sin{\left(x - 5 \right)}}{2} + \frac{\sin{\left(5 x - 5 \right)}}{10}+C$$

$$$\int \cos{\left(2 x \right)} \cos{\left(3 x - 5 \right)}\, dx = \left(\frac{\sin{\left(x - 5 \right)}}{2} + \frac{\sin{\left(5 x - 5 \right)}}{10}\right) + C$$$A