Integral of $$$\cos{\left(5 t \right)} \cos{\left(10 t \right)}$$$

Find $$$\int \cos{\left(5 t \right)} \cos{\left(10 t \right)}\, dt$$$.

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 $$$\alpha=5 t$$$ and $$$\beta=10 t$$$:

$${\color{red}{\int{\cos{\left(5 t \right)} \cos{\left(10 t \right)} d t}}} = {\color{red}{\int{\left(\frac{\cos{\left(5 t \right)}}{2} + \frac{\cos{\left(15 t \right)}}{2}\right)d t}}}$$

Apply the constant multiple rule $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(t \right)} = \cos{\left(5 t \right)} + \cos{\left(15 t \right)}$$$:

$${\color{red}{\int{\left(\frac{\cos{\left(5 t \right)}}{2} + \frac{\cos{\left(15 t \right)}}{2}\right)d t}}} = {\color{red}{\left(\frac{\int{\left(\cos{\left(5 t \right)} + \cos{\left(15 t \right)}\right)d t}}{2}\right)}}$$

Integrate term by term:

$$\frac{{\color{red}{\int{\left(\cos{\left(5 t \right)} + \cos{\left(15 t \right)}\right)d t}}}}{2} = \frac{{\color{red}{\left(\int{\cos{\left(5 t \right)} d t} + \int{\cos{\left(15 t \right)} d t}\right)}}}{2}$$

Let $$$u=5 t$$$.

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

The integral can be rewritten as

$$\frac{\int{\cos{\left(15 t \right)} d t}}{2} + \frac{{\color{red}{\int{\cos{\left(5 t \right)} d t}}}}{2} = \frac{\int{\cos{\left(15 t \right)} d t}}{2} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{5} d u}}}}{2}$$

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

$$\frac{\int{\cos{\left(15 t \right)} d t}}{2} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{5} d u}}}}{2} = \frac{\int{\cos{\left(15 t \right)} d t}}{2} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{5}\right)}}}{2}$$

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

$$\frac{\int{\cos{\left(15 t \right)} d t}}{2} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{10} = \frac{\int{\cos{\left(15 t \right)} d t}}{2} + \frac{{\color{red}{\sin{\left(u \right)}}}}{10}$$

Recall that $$$u=5 t$$$:

$$\frac{\int{\cos{\left(15 t \right)} d t}}{2} + \frac{\sin{\left({\color{red}{u}} \right)}}{10} = \frac{\int{\cos{\left(15 t \right)} d t}}{2} + \frac{\sin{\left({\color{red}{\left(5 t\right)}} \right)}}{10}$$

Let $$$u=15 t$$$.

Then $$$du=\left(15 t\right)^{\prime }dt = 15 dt$$$ (steps can be seen »), and we have that $$$dt = \frac{du}{15}$$$.

Thus,

$$\frac{\sin{\left(5 t \right)}}{10} + \frac{{\color{red}{\int{\cos{\left(15 t \right)} d t}}}}{2} = \frac{\sin{\left(5 t \right)}}{10} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{15} d u}}}}{2}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{15}$$$ and $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$$\frac{\sin{\left(5 t \right)}}{10} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{15} d u}}}}{2} = \frac{\sin{\left(5 t \right)}}{10} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{15}\right)}}}{2}$$

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

$$\frac{\sin{\left(5 t \right)}}{10} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{30} = \frac{\sin{\left(5 t \right)}}{10} + \frac{{\color{red}{\sin{\left(u \right)}}}}{30}$$

Recall that $$$u=15 t$$$:

$$\frac{\sin{\left(5 t \right)}}{10} + \frac{\sin{\left({\color{red}{u}} \right)}}{30} = \frac{\sin{\left(5 t \right)}}{10} + \frac{\sin{\left({\color{red}{\left(15 t\right)}} \right)}}{30}$$

Therefore,

$$\int{\cos{\left(5 t \right)} \cos{\left(10 t \right)} d t} = \frac{\sin{\left(5 t \right)}}{10} + \frac{\sin{\left(15 t \right)}}{30}$$

Add the constant of integration:

$$\int{\cos{\left(5 t \right)} \cos{\left(10 t \right)} d t} = \frac{\sin{\left(5 t \right)}}{10} + \frac{\sin{\left(15 t \right)}}{30}+C$$

$$$\int \cos{\left(5 t \right)} \cos{\left(10 t \right)}\, dt = \left(\frac{\sin{\left(5 t \right)}}{10} + \frac{\sin{\left(15 t \right)}}{30}\right) + C$$$A