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

The calculator will find the integral/antiderivative of $\sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}$, with steps shown.

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

### Solution

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 $dx = \frac{du}{4}$.

The integral becomes

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

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,

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

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