Integral of $$$a^{2} b^{2} \sin^{2}{\left(2 x \right)}$$$ with respect to $$$x$$$

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

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

$${\color{red}{\int{a^{2} b^{2} \sin^{2}{\left(2 x \right)} d x}}} = {\color{red}{\int{\frac{a^{2} b^{2} \left(1 - \cos{\left(4 x \right)}\right)}{2} d x}}}$$

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 $$$f{\left(x \right)} = a^{2} b^{2} \left(1 - \cos{\left(4 x \right)}\right)$$$:

$${\color{red}{\int{\frac{a^{2} b^{2} \left(1 - \cos{\left(4 x \right)}\right)}{2} d x}}} = {\color{red}{\left(\frac{\int{a^{2} b^{2} \left(1 - \cos{\left(4 x \right)}\right) d x}}{2}\right)}}$$

Expand the expression:

$$\frac{{\color{red}{\int{a^{2} b^{2} \left(1 - \cos{\left(4 x \right)}\right) d x}}}}{2} = \frac{{\color{red}{\int{\left(- a^{2} b^{2} \cos{\left(4 x \right)} + a^{2} b^{2}\right)d x}}}}{2}$$

Integrate term by term:

$$\frac{{\color{red}{\int{\left(- a^{2} b^{2} \cos{\left(4 x \right)} + a^{2} b^{2}\right)d x}}}}{2} = \frac{{\color{red}{\left(\int{a^{2} b^{2} d x} - \int{a^{2} b^{2} \cos{\left(4 x \right)} d x}\right)}}}{2}$$

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=a^{2} b^{2}$$$:

$$- \frac{\int{a^{2} b^{2} \cos{\left(4 x \right)} d x}}{2} + \frac{{\color{red}{\int{a^{2} b^{2} d x}}}}{2} = - \frac{\int{a^{2} b^{2} \cos{\left(4 x \right)} d x}}{2} + \frac{{\color{red}{a^{2} b^{2} x}}}{2}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=a^{2} b^{2}$$$ and $$$f{\left(x \right)} = \cos{\left(4 x \right)}$$$:

$$\frac{a^{2} b^{2} x}{2} - \frac{{\color{red}{\int{a^{2} b^{2} \cos{\left(4 x \right)} d x}}}}{2} = \frac{a^{2} b^{2} x}{2} - \frac{{\color{red}{a^{2} b^{2} \int{\cos{\left(4 x \right)} d x}}}}{2}$$

Let $$$u=4 x$$$.

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

The integral can be rewritten as

$$\frac{a^{2} b^{2} x}{2} - \frac{a^{2} b^{2} {\color{red}{\int{\cos{\left(4 x \right)} d x}}}}{2} = \frac{a^{2} b^{2} x}{2} - \frac{a^{2} b^{2} {\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{2}$$

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

$$\frac{a^{2} b^{2} x}{2} - \frac{a^{2} b^{2} {\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{2} = \frac{a^{2} b^{2} x}{2} - \frac{a^{2} b^{2} {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{4}\right)}}}{2}$$

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

$$\frac{a^{2} b^{2} x}{2} - \frac{a^{2} b^{2} {\color{red}{\int{\cos{\left(u \right)} d u}}}}{8} = \frac{a^{2} b^{2} x}{2} - \frac{a^{2} b^{2} {\color{red}{\sin{\left(u \right)}}}}{8}$$

Recall that $$$u=4 x$$$:

$$\frac{a^{2} b^{2} x}{2} - \frac{a^{2} b^{2} \sin{\left({\color{red}{u}} \right)}}{8} = \frac{a^{2} b^{2} x}{2} - \frac{a^{2} b^{2} \sin{\left({\color{red}{\left(4 x\right)}} \right)}}{8}$$

Therefore,

$$\int{a^{2} b^{2} \sin^{2}{\left(2 x \right)} d x} = \frac{a^{2} b^{2} x}{2} - \frac{a^{2} b^{2} \sin{\left(4 x \right)}}{8}$$

Simplify:

$$\int{a^{2} b^{2} \sin^{2}{\left(2 x \right)} d x} = \frac{a^{2} b^{2} \left(4 x - \sin{\left(4 x \right)}\right)}{8}$$

Add the constant of integration:

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

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