Integral of $$$\frac{\sin{\left(4 x \right)}}{\sin{\left(x \right)}}$$$

Find $$$\int \frac{\sin{\left(4 x \right)}}{\sin{\left(x \right)}}\, dx$$$.

Rewrite the integrand:

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

Simplify the integrand:

$${\color{red}{\int{\frac{- 8 \sin^{3}{\left(x \right)} \cos{\left(x \right)} + 4 \sin{\left(x \right)} \cos{\left(x \right)}}{\sin{\left(x \right)}} d x}}} = {\color{red}{\int{4 \left(1 - 2 \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)} d x}}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=4$$$ and $$$f{\left(x \right)} = \left(1 - 2 \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)}$$$:

$${\color{red}{\int{4 \left(1 - 2 \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)} d x}}} = {\color{red}{\left(4 \int{\left(1 - 2 \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)} d x}\right)}}$$

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

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

The integral can be rewritten as

$$4 {\color{red}{\int{\left(1 - 2 \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)} d x}}} = 4 {\color{red}{\int{\left(1 - 2 u^{2}\right)d u}}}$$

Integrate term by term:

$$4 {\color{red}{\int{\left(1 - 2 u^{2}\right)d u}}} = 4 {\color{red}{\left(\int{1 d u} - \int{2 u^{2} d u}\right)}}$$

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:

$$- 4 \int{2 u^{2} d u} + 4 {\color{red}{\int{1 d u}}} = - 4 \int{2 u^{2} d u} + 4 {\color{red}{u}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=2$$$ and $$$f{\left(u \right)} = u^{2}$$$:

$$4 u - 4 {\color{red}{\int{2 u^{2} d u}}} = 4 u - 4 {\color{red}{\left(2 \int{u^{2} d u}\right)}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=2$$$:

$$4 u - 8 {\color{red}{\int{u^{2} d u}}}=4 u - 8 {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=4 u - 8 {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

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

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

Therefore,

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

Simplify:

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

Add the constant of integration:

$$\int{\frac{\sin{\left(4 x \right)}}{\sin{\left(x \right)}} d x} = 2 \sin{\left(x \right)} + \frac{2 \sin{\left(3 x \right)}}{3}+C$$

$$$\int \frac{\sin{\left(4 x \right)}}{\sin{\left(x \right)}}\, dx = \left(2 \sin{\left(x \right)} + \frac{2 \sin{\left(3 x \right)}}{3}\right) + C$$$A