Integral of $$$\sin^{4}{\left(2 x \right)}$$$

Find $$$\int \sin^{4}{\left(2 x \right)}\, dx$$$.

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

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

The integral can be rewritten as

$${\color{red}{\int{\sin^{4}{\left(2 x \right)} d x}}} = {\color{red}{\int{\frac{\sin^{4}{\left(u \right)}}{2} d u}}}$$

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 $$$f{\left(u \right)} = \sin^{4}{\left(u \right)}$$$:

$${\color{red}{\int{\frac{\sin^{4}{\left(u \right)}}{2} d u}}} = {\color{red}{\left(\frac{\int{\sin^{4}{\left(u \right)} d u}}{2}\right)}}$$

Apply the power reducing formula $$$\sin^{4}{\left(\alpha \right)} = - \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{\cos{\left(4 \alpha \right)}}{8} + \frac{3}{8}$$$ with $$$\alpha= u $$$:

$$\frac{{\color{red}{\int{\sin^{4}{\left(u \right)} d u}}}}{2} = \frac{{\color{red}{\int{\left(- \frac{\cos{\left(2 u \right)}}{2} + \frac{\cos{\left(4 u \right)}}{8} + \frac{3}{8}\right)d u}}}}{2}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{8}$$$ and $$$f{\left(u \right)} = - 4 \cos{\left(2 u \right)} + \cos{\left(4 u \right)} + 3$$$:

$$\frac{{\color{red}{\int{\left(- \frac{\cos{\left(2 u \right)}}{2} + \frac{\cos{\left(4 u \right)}}{8} + \frac{3}{8}\right)d u}}}}{2} = \frac{{\color{red}{\left(\frac{\int{\left(- 4 \cos{\left(2 u \right)} + \cos{\left(4 u \right)} + 3\right)d u}}{8}\right)}}}{2}$$

Integrate term by term:

$$\frac{{\color{red}{\int{\left(- 4 \cos{\left(2 u \right)} + \cos{\left(4 u \right)} + 3\right)d u}}}}{16} = \frac{{\color{red}{\left(\int{3 d u} - \int{4 \cos{\left(2 u \right)} d u} + \int{\cos{\left(4 u \right)} d u}\right)}}}{16}$$

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=3$$$:

$$- \frac{\int{4 \cos{\left(2 u \right)} d u}}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{16} + \frac{{\color{red}{\int{3 d u}}}}{16} = - \frac{\int{4 \cos{\left(2 u \right)} d u}}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{16} + \frac{{\color{red}{\left(3 u\right)}}}{16}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=4$$$ and $$$f{\left(u \right)} = \cos{\left(2 u \right)}$$$:

$$\frac{3 u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{16} - \frac{{\color{red}{\int{4 \cos{\left(2 u \right)} d u}}}}{16} = \frac{3 u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{16} - \frac{{\color{red}{\left(4 \int{\cos{\left(2 u \right)} d u}\right)}}}{16}$$

Let $$$v=2 u$$$.

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

Thus,

$$\frac{3 u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{16} - \frac{{\color{red}{\int{\cos{\left(2 u \right)} d u}}}}{4} = \frac{3 u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{16} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{4}$$

Apply the constant multiple rule $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(v \right)} = \cos{\left(v \right)}$$$:

$$\frac{3 u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{16} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{4} = \frac{3 u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{16} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{2}\right)}}}{4}$$

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

$$\frac{3 u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{16} - \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{8} = \frac{3 u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{16} - \frac{{\color{red}{\sin{\left(v \right)}}}}{8}$$

Recall that $$$v=2 u$$$:

$$\frac{3 u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{16} - \frac{\sin{\left({\color{red}{v}} \right)}}{8} = \frac{3 u}{16} + \frac{\int{\cos{\left(4 u \right)} d u}}{16} - \frac{\sin{\left({\color{red}{\left(2 u\right)}} \right)}}{8}$$

Let $$$v=4 u$$$.

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

Thus,

$$\frac{3 u}{16} - \frac{\sin{\left(2 u \right)}}{8} + \frac{{\color{red}{\int{\cos{\left(4 u \right)} d u}}}}{16} = \frac{3 u}{16} - \frac{\sin{\left(2 u \right)}}{8} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}}{16}$$

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

$$\frac{3 u}{16} - \frac{\sin{\left(2 u \right)}}{8} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}}{16} = \frac{3 u}{16} - \frac{\sin{\left(2 u \right)}}{8} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{4}\right)}}}{16}$$

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

$$\frac{3 u}{16} - \frac{\sin{\left(2 u \right)}}{8} + \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{64} = \frac{3 u}{16} - \frac{\sin{\left(2 u \right)}}{8} + \frac{{\color{red}{\sin{\left(v \right)}}}}{64}$$

Recall that $$$v=4 u$$$:

$$\frac{3 u}{16} - \frac{\sin{\left(2 u \right)}}{8} + \frac{\sin{\left({\color{red}{v}} \right)}}{64} = \frac{3 u}{16} - \frac{\sin{\left(2 u \right)}}{8} + \frac{\sin{\left({\color{red}{\left(4 u\right)}} \right)}}{64}$$

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

$$- \frac{\sin{\left(2 {\color{red}{u}} \right)}}{8} + \frac{\sin{\left(4 {\color{red}{u}} \right)}}{64} + \frac{3 {\color{red}{u}}}{16} = - \frac{\sin{\left(2 {\color{red}{\left(2 x\right)}} \right)}}{8} + \frac{\sin{\left(4 {\color{red}{\left(2 x\right)}} \right)}}{64} + \frac{3 {\color{red}{\left(2 x\right)}}}{16}$$

Therefore,

$$\int{\sin^{4}{\left(2 x \right)} d x} = \frac{3 x}{8} - \frac{\sin{\left(4 x \right)}}{8} + \frac{\sin{\left(8 x \right)}}{64}$$

Simplify:

$$\int{\sin^{4}{\left(2 x \right)} d x} = \frac{24 x - 8 \sin{\left(4 x \right)} + \sin{\left(8 x \right)}}{64}$$

Add the constant of integration:

$$\int{\sin^{4}{\left(2 x \right)} d x} = \frac{24 x - 8 \sin{\left(4 x \right)} + \sin{\left(8 x \right)}}{64}+C$$

$$$\int \sin^{4}{\left(2 x \right)}\, dx = \frac{24 x - 8 \sin{\left(4 x \right)} + \sin{\left(8 x \right)}}{64} + C$$$A