Integral of $$$\sin{\left(14 x \right)} \cos{\left(9 x \right)}$$$
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Find $$$\int \sin{\left(14 x \right)} \cos{\left(9 x \right)}\, dx$$$.
Solution
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 $$$\alpha=14 x$$$ and $$$\beta=9 x$$$:
$${\color{red}{\int{\sin{\left(14 x \right)} \cos{\left(9 x \right)} d x}}} = {\color{red}{\int{\left(\frac{\sin{\left(5 x \right)}}{2} + \frac{\sin{\left(23 x \right)}}{2}\right)d x}}}$$
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 $$$f{\left(x \right)} = \sin{\left(5 x \right)} + \sin{\left(23 x \right)}$$$:
$${\color{red}{\int{\left(\frac{\sin{\left(5 x \right)}}{2} + \frac{\sin{\left(23 x \right)}}{2}\right)d x}}} = {\color{red}{\left(\frac{\int{\left(\sin{\left(5 x \right)} + \sin{\left(23 x \right)}\right)d x}}{2}\right)}}$$
Integrate term by term:
$$\frac{{\color{red}{\int{\left(\sin{\left(5 x \right)} + \sin{\left(23 x \right)}\right)d x}}}}{2} = \frac{{\color{red}{\left(\int{\sin{\left(5 x \right)} d x} + \int{\sin{\left(23 x \right)} d x}\right)}}}{2}$$
Let $$$u=5 x$$$.
Then $$$du=\left(5 x\right)^{\prime }dx = 5 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{5}$$$.
Thus,
$$\frac{\int{\sin{\left(23 x \right)} d x}}{2} + \frac{{\color{red}{\int{\sin{\left(5 x \right)} d x}}}}{2} = \frac{\int{\sin{\left(23 x \right)} d x}}{2} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{5} d u}}}}{2}$$
Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{5}$$$ and $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:
$$\frac{\int{\sin{\left(23 x \right)} d x}}{2} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{5} d u}}}}{2} = \frac{\int{\sin{\left(23 x \right)} d x}}{2} + \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{5}\right)}}}{2}$$
The integral of the sine is $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:
$$\frac{\int{\sin{\left(23 x \right)} d x}}{2} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{10} = \frac{\int{\sin{\left(23 x \right)} d x}}{2} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{10}$$
Recall that $$$u=5 x$$$:
$$\frac{\int{\sin{\left(23 x \right)} d x}}{2} - \frac{\cos{\left({\color{red}{u}} \right)}}{10} = \frac{\int{\sin{\left(23 x \right)} d x}}{2} - \frac{\cos{\left({\color{red}{\left(5 x\right)}} \right)}}{10}$$
Let $$$u=23 x$$$.
Then $$$du=\left(23 x\right)^{\prime }dx = 23 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{23}$$$.
Therefore,
$$- \frac{\cos{\left(5 x \right)}}{10} + \frac{{\color{red}{\int{\sin{\left(23 x \right)} d x}}}}{2} = - \frac{\cos{\left(5 x \right)}}{10} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{23} d u}}}}{2}$$
Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{23}$$$ and $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:
$$- \frac{\cos{\left(5 x \right)}}{10} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{23} d u}}}}{2} = - \frac{\cos{\left(5 x \right)}}{10} + \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{23}\right)}}}{2}$$
The integral of the sine is $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:
$$- \frac{\cos{\left(5 x \right)}}{10} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{46} = - \frac{\cos{\left(5 x \right)}}{10} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{46}$$
Recall that $$$u=23 x$$$:
$$- \frac{\cos{\left(5 x \right)}}{10} - \frac{\cos{\left({\color{red}{u}} \right)}}{46} = - \frac{\cos{\left(5 x \right)}}{10} - \frac{\cos{\left({\color{red}{\left(23 x\right)}} \right)}}{46}$$
Therefore,
$$\int{\sin{\left(14 x \right)} \cos{\left(9 x \right)} d x} = - \frac{\cos{\left(5 x \right)}}{10} - \frac{\cos{\left(23 x \right)}}{46}$$
Add the constant of integration:
$$\int{\sin{\left(14 x \right)} \cos{\left(9 x \right)} d x} = - \frac{\cos{\left(5 x \right)}}{10} - \frac{\cos{\left(23 x \right)}}{46}+C$$
Answer
$$$\int \sin{\left(14 x \right)} \cos{\left(9 x \right)}\, dx = \left(- \frac{\cos{\left(5 x \right)}}{10} - \frac{\cos{\left(23 x \right)}}{46}\right) + C$$$A