Integral of $$$\sin{\left(x \right)} \sin{\left(2 x \right)} \sin{\left(3 x \right)}$$$

Find $$$\int \sin{\left(x \right)} \sin{\left(2 x \right)} \sin{\left(3 x \right)}\, dx$$$.

Rewrite $$$\sin\left(x \right)\sin\left(2 x \right)$$$ using the formula $$$\sin\left(\alpha \right)\sin\left(\beta \right)=\frac{1}{2} \cos\left(\alpha-\beta \right)-\frac{1}{2} \cos\left(\alpha+\beta \right)$$$ with $$$\alpha=x$$$ and $$$\beta=2 x$$$:

$${\color{red}{\int{\sin{\left(x \right)} \sin{\left(2 x \right)} \sin{\left(3 x \right)} d x}}} = {\color{red}{\int{\left(\frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left(3 x \right)}}{2}\right) \sin{\left(3 x \right)} d x}}}$$

Expand the expression:

$${\color{red}{\int{\left(\frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left(3 x \right)}}{2}\right) \sin{\left(3 x \right)} d x}}} = {\color{red}{\int{\left(\frac{\sin{\left(3 x \right)} \cos{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)} \cos{\left(3 x \right)}}{2}\right)d x}}}$$

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 $$$f{\left(x \right)} = \sin{\left(3 x \right)} \cos{\left(x \right)} - \sin{\left(3 x \right)} \cos{\left(3 x \right)}$$$:

$${\color{red}{\int{\left(\frac{\sin{\left(3 x \right)} \cos{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)} \cos{\left(3 x \right)}}{2}\right)d x}}} = {\color{red}{\left(\frac{\int{\left(\sin{\left(3 x \right)} \cos{\left(x \right)} - \sin{\left(3 x \right)} \cos{\left(3 x \right)}\right)d x}}{2}\right)}}$$

Integrate term by term:

$$\frac{{\color{red}{\int{\left(\sin{\left(3 x \right)} \cos{\left(x \right)} - \sin{\left(3 x \right)} \cos{\left(3 x \right)}\right)d x}}}}{2} = \frac{{\color{red}{\left(\int{\sin{\left(3 x \right)} \cos{\left(x \right)} d x} - \int{\sin{\left(3 x \right)} \cos{\left(3 x \right)} d x}\right)}}}{2}$$

Rewrite the integrand using the formula $$$\sin\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \sin\left(\alpha-\beta \right)+\frac{1}{2} \sin\left(\alpha+\beta \right)$$$ with $$$\alpha=3 x$$$ and $$$\beta=x$$$:

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

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 $$$f{\left(x \right)} = \sin{\left(2 x \right)} + \sin{\left(4 x \right)}$$$:

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

Integrate term by term:

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

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

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

Therefore,

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

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

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

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

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

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

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

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

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

Thus,

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

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

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

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

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

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

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

Let $$$u=\sin{\left(3 x \right)}$$$.

Then $$$du=\left(\sin{\left(3 x \right)}\right)^{\prime }dx = 3 \cos{\left(3 x \right)} dx$$$ (steps can be seen »), and we have that $$$\cos{\left(3 x \right)} dx = \frac{du}{3}$$$.

Therefore,

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

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

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

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

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16} - \frac{{\color{red}{\int{u d u}}}}{6}=- \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16} - \frac{{\color{red}{\frac{u^{1 + 1}}{1 + 1}}}}{6}=- \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16} - \frac{{\color{red}{\left(\frac{u^{2}}{2}\right)}}}{6}$$

Recall that $$$u=\sin{\left(3 x \right)}$$$:

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16} - \frac{{\color{red}{u}}^{2}}{12} = - \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16} - \frac{{\color{red}{\sin{\left(3 x \right)}}}^{2}}{12}$$

Therefore,

$$\int{\sin{\left(x \right)} \sin{\left(2 x \right)} \sin{\left(3 x \right)} d x} = - \frac{\sin^{2}{\left(3 x \right)}}{12} - \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16}$$

Add the constant of integration:

$$\int{\sin{\left(x \right)} \sin{\left(2 x \right)} \sin{\left(3 x \right)} d x} = - \frac{\sin^{2}{\left(3 x \right)}}{12} - \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16}+C$$

$$$\int \sin{\left(x \right)} \sin{\left(2 x \right)} \sin{\left(3 x \right)}\, dx = \left(- \frac{\sin^{2}{\left(3 x \right)}}{12} - \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16}\right) + C$$$A