Integral of $$$4 \sin^{2}{\left(x \right)} \cos^{3}{\left(x \right)}$$$

Find $$$\int 4 \sin^{2}{\left(x \right)} \cos^{3}{\left(x \right)}\, dx$$$.

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

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

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

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

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{8}$$$ and $$$f{\left(x \right)} = 4 \left(1 - \cos{\left(2 x \right)}\right) \left(3 \cos{\left(x \right)} + \cos{\left(3 x \right)}\right)$$$:

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

Expand the expression:

$$\frac{{\color{red}{\int{4 \left(1 - \cos{\left(2 x \right)}\right) \left(3 \cos{\left(x \right)} + \cos{\left(3 x \right)}\right) d x}}}}{8} = \frac{{\color{red}{\int{\left(- 12 \cos{\left(x \right)} \cos{\left(2 x \right)} + 12 \cos{\left(x \right)} - 4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} + 4 \cos{\left(3 x \right)}\right)d x}}}}{8}$$

Integrate term by term:

$$\frac{{\color{red}{\int{\left(- 12 \cos{\left(x \right)} \cos{\left(2 x \right)} + 12 \cos{\left(x \right)} - 4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} + 4 \cos{\left(3 x \right)}\right)d x}}}}{8} = \frac{{\color{red}{\left(- \int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x} - \int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x} + \int{12 \cos{\left(x \right)} d x} + \int{4 \cos{\left(3 x \right)} d x}\right)}}}{8}$$

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

$$- \frac{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}{8} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} + \frac{\int{12 \cos{\left(x \right)} d x}}{8} + \frac{{\color{red}{\int{4 \cos{\left(3 x \right)} d x}}}}{8} = - \frac{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}{8} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} + \frac{\int{12 \cos{\left(x \right)} d x}}{8} + \frac{{\color{red}{\left(4 \int{\cos{\left(3 x \right)} d x}\right)}}}{8}$$

Let $$$u=3 x$$$.

Then $$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{3}$$$.

The integral becomes

$$- \frac{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}{8} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} + \frac{\int{12 \cos{\left(x \right)} d x}}{8} + \frac{{\color{red}{\int{\cos{\left(3 x \right)} d x}}}}{2} = - \frac{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}{8} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} + \frac{\int{12 \cos{\left(x \right)} d x}}{8} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{3} d u}}}}{2}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{3}$$$ and $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$$- \frac{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}{8} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} + \frac{\int{12 \cos{\left(x \right)} d x}}{8} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{3} d u}}}}{2} = - \frac{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}{8} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} + \frac{\int{12 \cos{\left(x \right)} d x}}{8} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{3}\right)}}}{2}$$

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

$$- \frac{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}{8} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} + \frac{\int{12 \cos{\left(x \right)} d x}}{8} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{6} = - \frac{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}{8} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} + \frac{\int{12 \cos{\left(x \right)} d x}}{8} + \frac{{\color{red}{\sin{\left(u \right)}}}}{6}$$

Recall that $$$u=3 x$$$:

$$- \frac{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}{8} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} + \frac{\int{12 \cos{\left(x \right)} d x}}{8} + \frac{\sin{\left({\color{red}{u}} \right)}}{6} = - \frac{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}{8} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} + \frac{\int{12 \cos{\left(x \right)} d x}}{8} + \frac{\sin{\left({\color{red}{\left(3 x\right)}} \right)}}{6}$$

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

$$\frac{\sin{\left(3 x \right)}}{6} - \frac{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}{8} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} + \frac{{\color{red}{\int{12 \cos{\left(x \right)} d x}}}}{8} = \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}{8} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} + \frac{{\color{red}{\left(12 \int{\cos{\left(x \right)} d x}\right)}}}{8}$$

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

$$\frac{\sin{\left(3 x \right)}}{6} - \frac{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}{8} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} + \frac{3 {\color{red}{\int{\cos{\left(x \right)} d x}}}}{2} = \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}{8} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} + \frac{3 {\color{red}{\sin{\left(x \right)}}}}{2}$$

Rewrite $$$\cos\left(x \right)\cos\left(2 x \right)$$$ 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 $$$\beta=2 x$$$:

$$\frac{3 \sin{\left(x \right)}}{2} + \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} - \frac{{\color{red}{\int{12 \cos{\left(x \right)} \cos{\left(2 x \right)} d x}}}}{8} = \frac{3 \sin{\left(x \right)}}{2} + \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} - \frac{{\color{red}{\int{\left(6 \cos{\left(x \right)} + 6 \cos{\left(3 x \right)}\right)d x}}}}{8}$$

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 $$$f{\left(x \right)} = 12 \cos{\left(x \right)} + 12 \cos{\left(3 x \right)}$$$:

$$\frac{3 \sin{\left(x \right)}}{2} + \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} - \frac{{\color{red}{\int{\left(6 \cos{\left(x \right)} + 6 \cos{\left(3 x \right)}\right)d x}}}}{8} = \frac{3 \sin{\left(x \right)}}{2} + \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} - \frac{{\color{red}{\left(\frac{\int{\left(12 \cos{\left(x \right)} + 12 \cos{\left(3 x \right)}\right)d x}}{2}\right)}}}{8}$$

Integrate term by term:

$$\frac{3 \sin{\left(x \right)}}{2} + \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} - \frac{{\color{red}{\int{\left(12 \cos{\left(x \right)} + 12 \cos{\left(3 x \right)}\right)d x}}}}{16} = \frac{3 \sin{\left(x \right)}}{2} + \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} - \frac{{\color{red}{\left(\int{12 \cos{\left(x \right)} d x} + \int{12 \cos{\left(3 x \right)} d x}\right)}}}{16}$$

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

$$\frac{3 \sin{\left(x \right)}}{2} + \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} - \frac{\int{12 \cos{\left(3 x \right)} d x}}{16} - \frac{{\color{red}{\int{12 \cos{\left(x \right)} d x}}}}{16} = \frac{3 \sin{\left(x \right)}}{2} + \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} - \frac{\int{12 \cos{\left(3 x \right)} d x}}{16} - \frac{{\color{red}{\left(12 \int{\cos{\left(x \right)} d x}\right)}}}{16}$$

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

$$\frac{3 \sin{\left(x \right)}}{2} + \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} - \frac{\int{12 \cos{\left(3 x \right)} d x}}{16} - \frac{3 {\color{red}{\int{\cos{\left(x \right)} d x}}}}{4} = \frac{3 \sin{\left(x \right)}}{2} + \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} - \frac{\int{12 \cos{\left(3 x \right)} d x}}{16} - \frac{3 {\color{red}{\sin{\left(x \right)}}}}{4}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=12$$$ and $$$f{\left(x \right)} = \cos{\left(3 x \right)}$$$:

$$\frac{3 \sin{\left(x \right)}}{4} + \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} - \frac{{\color{red}{\int{12 \cos{\left(3 x \right)} d x}}}}{16} = \frac{3 \sin{\left(x \right)}}{4} + \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} - \frac{{\color{red}{\left(12 \int{\cos{\left(3 x \right)} d x}\right)}}}{16}$$

The integral $$$\int{\cos{\left(3 x \right)} d x}$$$ was already calculated:

$$\int{\cos{\left(3 x \right)} d x} = \frac{\sin{\left(3 x \right)}}{3}$$

Therefore,

$$\frac{3 \sin{\left(x \right)}}{4} + \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} - \frac{3 {\color{red}{\int{\cos{\left(3 x \right)} d x}}}}{4} = \frac{3 \sin{\left(x \right)}}{4} + \frac{\sin{\left(3 x \right)}}{6} - \frac{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}{8} - \frac{3 {\color{red}{\left(\frac{\sin{\left(3 x \right)}}{3}\right)}}}{4}$$

Rewrite $$$\cos\left(2 x \right)\cos\left(3 x \right)$$$ 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 $$$\alpha=2 x$$$ and $$$\beta=3 x$$$:

$$\frac{3 \sin{\left(x \right)}}{4} - \frac{\sin{\left(3 x \right)}}{12} - \frac{{\color{red}{\int{4 \cos{\left(2 x \right)} \cos{\left(3 x \right)} d x}}}}{8} = \frac{3 \sin{\left(x \right)}}{4} - \frac{\sin{\left(3 x \right)}}{12} - \frac{{\color{red}{\int{\left(2 \cos{\left(x \right)} + 2 \cos{\left(5 x \right)}\right)d x}}}}{8}$$

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 $$$f{\left(x \right)} = 4 \cos{\left(x \right)} + 4 \cos{\left(5 x \right)}$$$:

$$\frac{3 \sin{\left(x \right)}}{4} - \frac{\sin{\left(3 x \right)}}{12} - \frac{{\color{red}{\int{\left(2 \cos{\left(x \right)} + 2 \cos{\left(5 x \right)}\right)d x}}}}{8} = \frac{3 \sin{\left(x \right)}}{4} - \frac{\sin{\left(3 x \right)}}{12} - \frac{{\color{red}{\left(\frac{\int{\left(4 \cos{\left(x \right)} + 4 \cos{\left(5 x \right)}\right)d x}}{2}\right)}}}{8}$$

Integrate term by term:

$$\frac{3 \sin{\left(x \right)}}{4} - \frac{\sin{\left(3 x \right)}}{12} - \frac{{\color{red}{\int{\left(4 \cos{\left(x \right)} + 4 \cos{\left(5 x \right)}\right)d x}}}}{16} = \frac{3 \sin{\left(x \right)}}{4} - \frac{\sin{\left(3 x \right)}}{12} - \frac{{\color{red}{\left(\int{4 \cos{\left(x \right)} d x} + \int{4 \cos{\left(5 x \right)} d x}\right)}}}{16}$$

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

$$\frac{3 \sin{\left(x \right)}}{4} - \frac{\sin{\left(3 x \right)}}{12} - \frac{\int{4 \cos{\left(5 x \right)} d x}}{16} - \frac{{\color{red}{\int{4 \cos{\left(x \right)} d x}}}}{16} = \frac{3 \sin{\left(x \right)}}{4} - \frac{\sin{\left(3 x \right)}}{12} - \frac{\int{4 \cos{\left(5 x \right)} d x}}{16} - \frac{{\color{red}{\left(4 \int{\cos{\left(x \right)} d x}\right)}}}{16}$$

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

$$\frac{3 \sin{\left(x \right)}}{4} - \frac{\sin{\left(3 x \right)}}{12} - \frac{\int{4 \cos{\left(5 x \right)} d x}}{16} - \frac{{\color{red}{\int{\cos{\left(x \right)} d x}}}}{4} = \frac{3 \sin{\left(x \right)}}{4} - \frac{\sin{\left(3 x \right)}}{12} - \frac{\int{4 \cos{\left(5 x \right)} d x}}{16} - \frac{{\color{red}{\sin{\left(x \right)}}}}{4}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=4$$$ and $$$f{\left(x \right)} = \cos{\left(5 x \right)}$$$:

$$\frac{\sin{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)}}{12} - \frac{{\color{red}{\int{4 \cos{\left(5 x \right)} d x}}}}{16} = \frac{\sin{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)}}{12} - \frac{{\color{red}{\left(4 \int{\cos{\left(5 x \right)} d x}\right)}}}{16}$$

Let $$$v=5 x$$$.

Then $$$dv=\left(5 x\right)^{\prime }dx = 5 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{dv}{5}$$$.

The integral becomes

$$\frac{\sin{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)}}{12} - \frac{{\color{red}{\int{\cos{\left(5 x \right)} d x}}}}{4} = \frac{\sin{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)}}{12} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{5} d v}}}}{4}$$

Apply the constant multiple rule $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ with $$$c=\frac{1}{5}$$$ and $$$f{\left(v \right)} = \cos{\left(v \right)}$$$:

$$\frac{\sin{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)}}{12} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{5} d v}}}}{4} = \frac{\sin{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)}}{12} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{5}\right)}}}{4}$$

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

$$\frac{\sin{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)}}{12} - \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{20} = \frac{\sin{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)}}{12} - \frac{{\color{red}{\sin{\left(v \right)}}}}{20}$$

Recall that $$$v=5 x$$$:

$$\frac{\sin{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)}}{12} - \frac{\sin{\left({\color{red}{v}} \right)}}{20} = \frac{\sin{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)}}{12} - \frac{\sin{\left({\color{red}{\left(5 x\right)}} \right)}}{20}$$

Therefore,

$$\int{4 \sin^{2}{\left(x \right)} \cos^{3}{\left(x \right)} d x} = \frac{\sin{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)}}{12} - \frac{\sin{\left(5 x \right)}}{20}$$

Add the constant of integration:

$$\int{4 \sin^{2}{\left(x \right)} \cos^{3}{\left(x \right)} d x} = \frac{\sin{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)}}{12} - \frac{\sin{\left(5 x \right)}}{20}+C$$

$$$\int 4 \sin^{2}{\left(x \right)} \cos^{3}{\left(x \right)}\, dx = \left(\frac{\sin{\left(x \right)}}{2} - \frac{\sin{\left(3 x \right)}}{12} - \frac{\sin{\left(5 x \right)}}{20}\right) + C$$$A