Integral of $$$\frac{\sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{916}$$$

Find $$$\int \frac{\sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{916}\, dx$$$.

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 $$$\beta=2 x$$$:

$${\color{red}{\int{\frac{\sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{916} d x}}} = {\color{red}{\int{\frac{\left(\frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left(3 x \right)}}{2}\right) \cos{\left(x \right)}}{916} d x}}}$$

Expand the expression:

$${\color{red}{\int{\frac{\left(\frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left(3 x \right)}}{2}\right) \cos{\left(x \right)}}{916} d x}}} = {\color{red}{\int{\left(\frac{\cos^{2}{\left(x \right)}}{1832} - \frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{1832}\right)d x}}}$$

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 $$$f{\left(x \right)} = \frac{\cos^{2}{\left(x \right)}}{916} - \frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916}$$$:

$${\color{red}{\int{\left(\frac{\cos^{2}{\left(x \right)}}{1832} - \frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{1832}\right)d x}}} = {\color{red}{\left(\frac{\int{\left(\frac{\cos^{2}{\left(x \right)}}{916} - \frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916}\right)d x}}{2}\right)}}$$

Integrate term by term:

$$\frac{{\color{red}{\int{\left(\frac{\cos^{2}{\left(x \right)}}{916} - \frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916}\right)d x}}}}{2} = \frac{{\color{red}{\left(- \int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x} + \int{\frac{\cos^{2}{\left(x \right)}}{916} d x}\right)}}}{2}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{916}$$$ and $$$f{\left(x \right)} = \cos^{2}{\left(x \right)}$$$:

$$- \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{{\color{red}{\int{\frac{\cos^{2}{\left(x \right)}}{916} d x}}}}{2} = - \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{{\color{red}{\left(\frac{\int{\cos^{2}{\left(x \right)} d x}}{916}\right)}}}{2}$$

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

$$- \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{{\color{red}{\int{\cos^{2}{\left(x \right)} d x}}}}{1832} = - \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)d x}}}}{1832}$$

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

$$- \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)d x}}}}{1832} = - \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{{\color{red}{\left(\frac{\int{\left(\cos{\left(2 x \right)} + 1\right)d x}}{2}\right)}}}{1832}$$

Integrate term by term:

$$- \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{{\color{red}{\int{\left(\cos{\left(2 x \right)} + 1\right)d x}}}}{3664} = - \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{{\color{red}{\left(\int{1 d x} + \int{\cos{\left(2 x \right)} d x}\right)}}}{3664}$$

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

$$- \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{\int{\cos{\left(2 x \right)} d x}}{3664} + \frac{{\color{red}{\int{1 d x}}}}{3664} = - \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{\int{\cos{\left(2 x \right)} d x}}{3664} + \frac{{\color{red}{x}}}{3664}$$

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

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

Thus,

$$\frac{x}{3664} - \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{{\color{red}{\int{\cos{\left(2 x \right)} d x}}}}{3664} = \frac{x}{3664} - \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{3664}$$

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

$$\frac{x}{3664} - \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{3664} = \frac{x}{3664} - \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}}{3664}$$

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

$$\frac{x}{3664} - \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{7328} = \frac{x}{3664} - \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{{\color{red}{\sin{\left(u \right)}}}}{7328}$$

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

$$\frac{x}{3664} - \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{\sin{\left({\color{red}{u}} \right)}}{7328} = \frac{x}{3664} - \frac{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}{2} + \frac{\sin{\left({\color{red}{\left(2 x\right)}} \right)}}{7328}$$

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

$$\frac{x}{3664} + \frac{\sin{\left(2 x \right)}}{7328} - \frac{{\color{red}{\int{\frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{916} d x}}}}{2} = \frac{x}{3664} + \frac{\sin{\left(2 x \right)}}{7328} - \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{1832} + \frac{\cos{\left(4 x \right)}}{1832}\right)d x}}}}{2}$$

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 $$$f{\left(x \right)} = \frac{\cos{\left(2 x \right)}}{916} + \frac{\cos{\left(4 x \right)}}{916}$$$:

$$\frac{x}{3664} + \frac{\sin{\left(2 x \right)}}{7328} - \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{1832} + \frac{\cos{\left(4 x \right)}}{1832}\right)d x}}}}{2} = \frac{x}{3664} + \frac{\sin{\left(2 x \right)}}{7328} - \frac{{\color{red}{\left(\frac{\int{\left(\frac{\cos{\left(2 x \right)}}{916} + \frac{\cos{\left(4 x \right)}}{916}\right)d x}}{2}\right)}}}{2}$$

Integrate term by term:

$$\frac{x}{3664} + \frac{\sin{\left(2 x \right)}}{7328} - \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{916} + \frac{\cos{\left(4 x \right)}}{916}\right)d x}}}}{4} = \frac{x}{3664} + \frac{\sin{\left(2 x \right)}}{7328} - \frac{{\color{red}{\left(\int{\frac{\cos{\left(2 x \right)}}{916} d x} + \int{\frac{\cos{\left(4 x \right)}}{916} d x}\right)}}}{4}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{916}$$$ and $$$f{\left(x \right)} = \cos{\left(2 x \right)}$$$:

$$\frac{x}{3664} + \frac{\sin{\left(2 x \right)}}{7328} - \frac{\int{\frac{\cos{\left(4 x \right)}}{916} d x}}{4} - \frac{{\color{red}{\int{\frac{\cos{\left(2 x \right)}}{916} d x}}}}{4} = \frac{x}{3664} + \frac{\sin{\left(2 x \right)}}{7328} - \frac{\int{\frac{\cos{\left(4 x \right)}}{916} d x}}{4} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(2 x \right)} d x}}{916}\right)}}}{4}$$

The integral $$$\int{\cos{\left(2 x \right)} d x}$$$ was already calculated:

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

Therefore,

$$\frac{x}{3664} + \frac{\sin{\left(2 x \right)}}{7328} - \frac{\int{\frac{\cos{\left(4 x \right)}}{916} d x}}{4} - \frac{{\color{red}{\int{\cos{\left(2 x \right)} d x}}}}{3664} = \frac{x}{3664} + \frac{\sin{\left(2 x \right)}}{7328} - \frac{\int{\frac{\cos{\left(4 x \right)}}{916} d x}}{4} - \frac{{\color{red}{\left(\frac{\sin{\left(2 x \right)}}{2}\right)}}}{3664}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{916}$$$ and $$$f{\left(x \right)} = \cos{\left(4 x \right)}$$$:

$$\frac{x}{3664} - \frac{{\color{red}{\int{\frac{\cos{\left(4 x \right)}}{916} d x}}}}{4} = \frac{x}{3664} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(4 x \right)} d x}}{916}\right)}}}{4}$$

Let $$$v=4 x$$$.

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

The integral can be rewritten as

$$\frac{x}{3664} - \frac{{\color{red}{\int{\cos{\left(4 x \right)} d x}}}}{3664} = \frac{x}{3664} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}}{3664}$$

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

$$\frac{x}{3664} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}}{3664} = \frac{x}{3664} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{4}\right)}}}{3664}$$

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

$$\frac{x}{3664} - \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{14656} = \frac{x}{3664} - \frac{{\color{red}{\sin{\left(v \right)}}}}{14656}$$

Recall that $$$v=4 x$$$:

$$\frac{x}{3664} - \frac{\sin{\left({\color{red}{v}} \right)}}{14656} = \frac{x}{3664} - \frac{\sin{\left({\color{red}{\left(4 x\right)}} \right)}}{14656}$$

Therefore,

$$\int{\frac{\sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{916} d x} = \frac{x}{3664} - \frac{\sin{\left(4 x \right)}}{14656}$$

Add the constant of integration:

$$\int{\frac{\sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{916} d x} = \frac{x}{3664} - \frac{\sin{\left(4 x \right)}}{14656}+C$$

$$$\int \frac{\sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{916}\, dx = \left(\frac{x}{3664} - \frac{\sin{\left(4 x \right)}}{14656}\right) + C$$$A