Integral of $$$\cos{\left(4 \theta \right)}$$$

Find $$$\int \cos{\left(4 \theta \right)}\, d\theta$$$.

Let $$$u=4 \theta$$$.

Then $$$du=\left(4 \theta\right)^{\prime }d\theta = 4 d\theta$$$ (steps can be seen »), and we have that $$$d\theta = \frac{du}{4}$$$.

So,

$${\color{red}{\int{\cos{\left(4 \theta \right)} d \theta}}} = {\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{4}$$$ and $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$${\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}} = {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{4}\right)}}$$

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

$$\frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{4} = \frac{{\color{red}{\sin{\left(u \right)}}}}{4}$$

Recall that $$$u=4 \theta$$$:

$$\frac{\sin{\left({\color{red}{u}} \right)}}{4} = \frac{\sin{\left({\color{red}{\left(4 \theta\right)}} \right)}}{4}$$

Therefore,

$$\int{\cos{\left(4 \theta \right)} d \theta} = \frac{\sin{\left(4 \theta \right)}}{4}$$

Add the constant of integration:

$$\int{\cos{\left(4 \theta \right)} d \theta} = \frac{\sin{\left(4 \theta \right)}}{4}+C$$

$$$\int \cos{\left(4 \theta \right)}\, d\theta = \frac{\sin{\left(4 \theta \right)}}{4} + C$$$A