Integral of $\sin{\left(14 x \right)} \cos{\left(5 x \right)}$

Find $\int \sin{\left(14 x \right)} \cos{\left(5 x \right)}\, dx$.

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 $\beta=5 x$:

$${\color{red}{\int{\sin{\left(14 x \right)} \cos{\left(5 x \right)} d x}}} = {\color{red}{\int{\left(\frac{\sin{\left(9 x \right)}}{2} + \frac{\sin{\left(19 x \right)}}{2}\right)d x}}}$$

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 $f{\left(x \right)} = \sin{\left(9 x \right)} + \sin{\left(19 x \right)}$:

$${\color{red}{\int{\left(\frac{\sin{\left(9 x \right)}}{2} + \frac{\sin{\left(19 x \right)}}{2}\right)d x}}} = {\color{red}{\left(\frac{\int{\left(\sin{\left(9 x \right)} + \sin{\left(19 x \right)}\right)d x}}{2}\right)}}$$

Integrate term by term:

$$\frac{{\color{red}{\int{\left(\sin{\left(9 x \right)} + \sin{\left(19 x \right)}\right)d x}}}}{2} = \frac{{\color{red}{\left(\int{\sin{\left(9 x \right)} d x} + \int{\sin{\left(19 x \right)} d x}\right)}}}{2}$$

Let $u=9 x$.

Then $du=\left(9 x\right)^{\prime }dx = 9 dx$ (steps can be seen »), and we have that $dx = \frac{du}{9}$.

Therefore,

$$\frac{\int{\sin{\left(19 x \right)} d x}}{2} + \frac{{\color{red}{\int{\sin{\left(9 x \right)} d x}}}}{2} = \frac{\int{\sin{\left(19 x \right)} d x}}{2} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{9} d u}}}}{2}$$

Apply the constant multiple rule $\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$ with $c=\frac{1}{9}$ and $f{\left(u \right)} = \sin{\left(u \right)}$:

$$\frac{\int{\sin{\left(19 x \right)} d x}}{2} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{9} d u}}}}{2} = \frac{\int{\sin{\left(19 x \right)} d x}}{2} + \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{9}\right)}}}{2}$$

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

$$\frac{\int{\sin{\left(19 x \right)} d x}}{2} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{18} = \frac{\int{\sin{\left(19 x \right)} d x}}{2} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{18}$$

Recall that $u=9 x$:

$$\frac{\int{\sin{\left(19 x \right)} d x}}{2} - \frac{\cos{\left({\color{red}{u}} \right)}}{18} = \frac{\int{\sin{\left(19 x \right)} d x}}{2} - \frac{\cos{\left({\color{red}{\left(9 x\right)}} \right)}}{18}$$

Let $u=19 x$.

Then $du=\left(19 x\right)^{\prime }dx = 19 dx$ (steps can be seen »), and we have that $dx = \frac{du}{19}$.

Thus,

$$- \frac{\cos{\left(9 x \right)}}{18} + \frac{{\color{red}{\int{\sin{\left(19 x \right)} d x}}}}{2} = - \frac{\cos{\left(9 x \right)}}{18} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{19} d u}}}}{2}$$

Apply the constant multiple rule $\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$ with $c=\frac{1}{19}$ and $f{\left(u \right)} = \sin{\left(u \right)}$:

$$- \frac{\cos{\left(9 x \right)}}{18} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{19} d u}}}}{2} = - \frac{\cos{\left(9 x \right)}}{18} + \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{19}\right)}}}{2}$$

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

$$- \frac{\cos{\left(9 x \right)}}{18} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{38} = - \frac{\cos{\left(9 x \right)}}{18} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{38}$$

Recall that $u=19 x$:

$$- \frac{\cos{\left(9 x \right)}}{18} - \frac{\cos{\left({\color{red}{u}} \right)}}{38} = - \frac{\cos{\left(9 x \right)}}{18} - \frac{\cos{\left({\color{red}{\left(19 x\right)}} \right)}}{38}$$

Therefore,

$$\int{\sin{\left(14 x \right)} \cos{\left(5 x \right)} d x} = - \frac{\cos{\left(9 x \right)}}{18} - \frac{\cos{\left(19 x \right)}}{38}$$

Add the constant of integration:

$$\int{\sin{\left(14 x \right)} \cos{\left(5 x \right)} d x} = - \frac{\cos{\left(9 x \right)}}{18} - \frac{\cos{\left(19 x \right)}}{38}+C$$

$\int \sin{\left(14 x \right)} \cos{\left(5 x \right)}\, dx = \left(- \frac{\cos{\left(9 x \right)}}{18} - \frac{\cos{\left(19 x \right)}}{38}\right) + C$A