Integral of $$$4 \sin^{3}{\left(7 x \right)} \cos{\left(7 x \right)}$$$

Find $$$\int 4 \sin^{3}{\left(7 x \right)} \cos{\left(7 x \right)}\, dx$$$.

Apply the power reducing formula $$$\sin^{3}{\left(\alpha \right)} = \frac{3 \sin{\left(\alpha \right)}}{4} - \frac{\sin{\left(3 \alpha \right)}}{4}$$$ with $$$\alpha=7 x$$$:

$${\color{red}{\int{4 \sin^{3}{\left(7 x \right)} \cos{\left(7 x \right)} d x}}} = {\color{red}{\int{\left(3 \sin{\left(7 x \right)} - \sin{\left(21 x \right)}\right) \cos{\left(7 x \right)} d x}}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{4}$$$ and $$$f{\left(x \right)} = 4 \left(3 \sin{\left(7 x \right)} - \sin{\left(21 x \right)}\right) \cos{\left(7 x \right)}$$$:

$${\color{red}{\int{\left(3 \sin{\left(7 x \right)} - \sin{\left(21 x \right)}\right) \cos{\left(7 x \right)} d x}}} = {\color{red}{\left(\frac{\int{4 \left(3 \sin{\left(7 x \right)} - \sin{\left(21 x \right)}\right) \cos{\left(7 x \right)} d x}}{4}\right)}}$$

Expand the expression:

$$\frac{{\color{red}{\int{4 \left(3 \sin{\left(7 x \right)} - \sin{\left(21 x \right)}\right) \cos{\left(7 x \right)} d x}}}}{4} = \frac{{\color{red}{\int{\left(12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} - 4 \sin{\left(21 x \right)} \cos{\left(7 x \right)}\right)d x}}}}{4}$$

Integrate term by term:

$$\frac{{\color{red}{\int{\left(12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} - 4 \sin{\left(21 x \right)} \cos{\left(7 x \right)}\right)d x}}}}{4} = \frac{{\color{red}{\left(\int{12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x} - \int{4 \sin{\left(21 x \right)} \cos{\left(7 x \right)} d x}\right)}}}{4}$$

Rewrite $$$\sin\left(21 x \right)\cos\left(7 x \right)$$$ 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 $$$\alpha=21 x$$$ and $$$\beta=7 x$$$:

$$\frac{\int{12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}{4} - \frac{{\color{red}{\int{4 \sin{\left(21 x \right)} \cos{\left(7 x \right)} d x}}}}{4} = \frac{\int{12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}{4} - \frac{{\color{red}{\int{\left(2 \sin{\left(14 x \right)} + 2 \sin{\left(28 x \right)}\right)d x}}}}{4}$$

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 $$$f{\left(x \right)} = 4 \sin{\left(14 x \right)} + 4 \sin{\left(28 x \right)}$$$:

$$\frac{\int{12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}{4} - \frac{{\color{red}{\int{\left(2 \sin{\left(14 x \right)} + 2 \sin{\left(28 x \right)}\right)d x}}}}{4} = \frac{\int{12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}{4} - \frac{{\color{red}{\left(\frac{\int{\left(4 \sin{\left(14 x \right)} + 4 \sin{\left(28 x \right)}\right)d x}}{2}\right)}}}{4}$$

Integrate term by term:

$$\frac{\int{12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}{4} - \frac{{\color{red}{\int{\left(4 \sin{\left(14 x \right)} + 4 \sin{\left(28 x \right)}\right)d x}}}}{8} = \frac{\int{12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}{4} - \frac{{\color{red}{\left(\int{4 \sin{\left(14 x \right)} d x} + \int{4 \sin{\left(28 x \right)} d x}\right)}}}{8}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=4$$$ and $$$f{\left(x \right)} = \sin{\left(14 x \right)}$$$:

$$\frac{\int{12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}{4} - \frac{\int{4 \sin{\left(28 x \right)} d x}}{8} - \frac{{\color{red}{\int{4 \sin{\left(14 x \right)} d x}}}}{8} = \frac{\int{12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}{4} - \frac{\int{4 \sin{\left(28 x \right)} d x}}{8} - \frac{{\color{red}{\left(4 \int{\sin{\left(14 x \right)} d x}\right)}}}{8}$$

Let $$$u=14 x$$$.

Then $$$du=\left(14 x\right)^{\prime }dx = 14 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{14}$$$.

The integral becomes

$$\frac{\int{12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}{4} - \frac{\int{4 \sin{\left(28 x \right)} d x}}{8} - \frac{{\color{red}{\int{\sin{\left(14 x \right)} d x}}}}{2} = \frac{\int{12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}{4} - \frac{\int{4 \sin{\left(28 x \right)} d x}}{8} - \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{14} d u}}}}{2}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{14}$$$ and $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$$\frac{\int{12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}{4} - \frac{\int{4 \sin{\left(28 x \right)} d x}}{8} - \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{14} d u}}}}{2} = \frac{\int{12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}{4} - \frac{\int{4 \sin{\left(28 x \right)} d x}}{8} - \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{14}\right)}}}{2}$$

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

$$\frac{\int{12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}{4} - \frac{\int{4 \sin{\left(28 x \right)} d x}}{8} - \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{28} = \frac{\int{12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}{4} - \frac{\int{4 \sin{\left(28 x \right)} d x}}{8} - \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{28}$$

Recall that $$$u=14 x$$$:

$$\frac{\int{12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}{4} - \frac{\int{4 \sin{\left(28 x \right)} d x}}{8} + \frac{\cos{\left({\color{red}{u}} \right)}}{28} = \frac{\int{12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}{4} - \frac{\int{4 \sin{\left(28 x \right)} d x}}{8} + \frac{\cos{\left({\color{red}{\left(14 x\right)}} \right)}}{28}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=4$$$ and $$$f{\left(x \right)} = \sin{\left(28 x \right)}$$$:

$$\frac{\cos{\left(14 x \right)}}{28} + \frac{\int{12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}{4} - \frac{{\color{red}{\int{4 \sin{\left(28 x \right)} d x}}}}{8} = \frac{\cos{\left(14 x \right)}}{28} + \frac{\int{12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}{4} - \frac{{\color{red}{\left(4 \int{\sin{\left(28 x \right)} d x}\right)}}}{8}$$

Let $$$u=28 x$$$.

Then $$$du=\left(28 x\right)^{\prime }dx = 28 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{28}$$$.

So,

$$\frac{\cos{\left(14 x \right)}}{28} + \frac{\int{12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}{4} - \frac{{\color{red}{\int{\sin{\left(28 x \right)} d x}}}}{2} = \frac{\cos{\left(14 x \right)}}{28} + \frac{\int{12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}{4} - \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{28} d u}}}}{2}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{28}$$$ and $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$$\frac{\cos{\left(14 x \right)}}{28} + \frac{\int{12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}{4} - \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{28} d u}}}}{2} = \frac{\cos{\left(14 x \right)}}{28} + \frac{\int{12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}{4} - \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{28}\right)}}}{2}$$

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

$$\frac{\cos{\left(14 x \right)}}{28} + \frac{\int{12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}{4} - \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{56} = \frac{\cos{\left(14 x \right)}}{28} + \frac{\int{12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}{4} - \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{56}$$

Recall that $$$u=28 x$$$:

$$\frac{\cos{\left(14 x \right)}}{28} + \frac{\int{12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}{4} + \frac{\cos{\left({\color{red}{u}} \right)}}{56} = \frac{\cos{\left(14 x \right)}}{28} + \frac{\int{12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}{4} + \frac{\cos{\left({\color{red}{\left(28 x\right)}} \right)}}{56}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=12$$$ and $$$f{\left(x \right)} = \sin{\left(7 x \right)} \cos{\left(7 x \right)}$$$:

$$\frac{\cos{\left(14 x \right)}}{28} + \frac{\cos{\left(28 x \right)}}{56} + \frac{{\color{red}{\int{12 \sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}}}{4} = \frac{\cos{\left(14 x \right)}}{28} + \frac{\cos{\left(28 x \right)}}{56} + \frac{{\color{red}{\left(12 \int{\sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}\right)}}}{4}$$

Let $$$u=\sin{\left(7 x \right)}$$$.

Then $$$du=\left(\sin{\left(7 x \right)}\right)^{\prime }dx = 7 \cos{\left(7 x \right)} dx$$$ (steps can be seen »), and we have that $$$\cos{\left(7 x \right)} dx = \frac{du}{7}$$$.

The integral becomes

$$\frac{\cos{\left(14 x \right)}}{28} + \frac{\cos{\left(28 x \right)}}{56} + 3 {\color{red}{\int{\sin{\left(7 x \right)} \cos{\left(7 x \right)} d x}}} = \frac{\cos{\left(14 x \right)}}{28} + \frac{\cos{\left(28 x \right)}}{56} + 3 {\color{red}{\int{\frac{u}{7} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{7}$$$ and $$$f{\left(u \right)} = u$$$:

$$\frac{\cos{\left(14 x \right)}}{28} + \frac{\cos{\left(28 x \right)}}{56} + 3 {\color{red}{\int{\frac{u}{7} d u}}} = \frac{\cos{\left(14 x \right)}}{28} + \frac{\cos{\left(28 x \right)}}{56} + 3 {\color{red}{\left(\frac{\int{u d u}}{7}\right)}}$$

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

$$\frac{\cos{\left(14 x \right)}}{28} + \frac{\cos{\left(28 x \right)}}{56} + \frac{3 {\color{red}{\int{u d u}}}}{7}=\frac{\cos{\left(14 x \right)}}{28} + \frac{\cos{\left(28 x \right)}}{56} + \frac{3 {\color{red}{\frac{u^{1 + 1}}{1 + 1}}}}{7}=\frac{\cos{\left(14 x \right)}}{28} + \frac{\cos{\left(28 x \right)}}{56} + \frac{3 {\color{red}{\left(\frac{u^{2}}{2}\right)}}}{7}$$

Recall that $$$u=\sin{\left(7 x \right)}$$$:

$$\frac{\cos{\left(14 x \right)}}{28} + \frac{\cos{\left(28 x \right)}}{56} + \frac{3 {\color{red}{u}}^{2}}{14} = \frac{\cos{\left(14 x \right)}}{28} + \frac{\cos{\left(28 x \right)}}{56} + \frac{3 {\color{red}{\sin{\left(7 x \right)}}}^{2}}{14}$$

Therefore,

$$\int{4 \sin^{3}{\left(7 x \right)} \cos{\left(7 x \right)} d x} = \frac{3 \sin^{2}{\left(7 x \right)}}{14} + \frac{\cos{\left(14 x \right)}}{28} + \frac{\cos{\left(28 x \right)}}{56}$$

Simplify:

$$\int{4 \sin^{3}{\left(7 x \right)} \cos{\left(7 x \right)} d x} = \frac{\sin^{4}{\left(7 x \right)}}{7} + \frac{3}{56}$$

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

$$\int{4 \sin^{3}{\left(7 x \right)} \cos{\left(7 x \right)} d x} = \frac{\sin^{4}{\left(7 x \right)}}{7}+C$$

$$$\int 4 \sin^{3}{\left(7 x \right)} \cos{\left(7 x \right)}\, dx = \frac{\sin^{4}{\left(7 x \right)}}{7} + C$$$A