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

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

Rewrite the integrand using the double angle formula $$$\sin\left(2 x \right)\cos\left(2 x \right)=\frac{1}{2}\sin\left( 2 x \right)$$$:

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

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{4}$$$ and $$$f{\left(x \right)} = \sin^{2}{\left(2 x \right)}$$$:

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

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

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

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

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

Integrate term by term:

$$\frac{\color{red}{\int{\left(1 - \cos{\left(4 x \right)}\right)d x}}}{8} = \frac{\color{red}{\left(\int{1 d x} - \int{\cos{\left(4 x \right)} d x}\right)}}{8}$$

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=1$$$:

$$- \frac{\int{\cos{\left(4 x \right)} d x}}{8} + \frac{\color{red}{\int{1 d x}}}{8} = - \frac{\int{\cos{\left(4 x \right)} d x}}{8} + \frac{\color{red}{x}}{8}$$

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

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

Thus,

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

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

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

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

$$\frac{x}{8} - \frac{\color{red}{\int{\cos{\left(u \right)} d u}}}{32} = \frac{x}{8} - \frac{\color{red}{\sin{\left(u \right)}}}{32}$$

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

$$\frac{x}{8} - \frac{\sin{\left(\color{red}{u} \right)}}{32} = \frac{x}{8} - \frac{\sin{\left(\color{red}{\left(4 x\right)} \right)}}{32}$$

Therefore,

$$\int{\sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)} d x} = \frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32}$$

Add the constant of integration:

$$\int{\sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)} d x} = \frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32}+C$$

Answer: $$$\int{\sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)} d x}=\frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32}+C$$$