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

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

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

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

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 $$$f{\left(x \right)} = \left(1 - \cos{\left(2 x \right)}\right) \sin{\left(2 x \right)}$$$:

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

Expand the expression:

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

Integrate term by term:

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

Let $$$u=\sin{\left(2 x \right)}$$$.

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

So,

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

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

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

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=1$$$:

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

Recall that $$$u=\sin{\left(2 x \right)}$$$:

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

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

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

So,

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

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

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

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

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

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

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

Therefore,

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

Simplify:

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

Add the constant of integration (and remove the constant from the expression):

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

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