Integral of $$$\frac{b^{2} \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{a^{2}}$$$ with respect to $$$x$$$

Find $$$\int \frac{b^{2} \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{a^{2}}\, dx$$$.

Rewrite the integrand:

$${\color{red}{\int{\frac{b^{2} \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{a^{2}} d x}}} = {\color{red}{\int{\frac{2 b^{2} \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}}{a^{2}} d x}}}$$

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

$${\color{red}{\int{\frac{2 b^{2} \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}}{a^{2}} d x}}} = {\color{red}{\left(\frac{2 b^{2} \int{\sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)} d x}}{a^{2}}\right)}}$$

Rewrite the integrand using the double angle formula $$$\sin\left(x \right)\cos\left(x \right)=\frac{1}{2}\sin\left( 2 x \right)$$$:

$$\frac{2 b^{2} {\color{red}{\int{\sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)} d x}}}}{a^{2}} = \frac{2 b^{2} {\color{red}{\int{\frac{\sin^{2}{\left(2 x \right)}}{4} d x}}}}{a^{2}}$$

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 $$$f{\left(x \right)} = \sin^{2}{\left(2 x \right)}$$$:

$$\frac{2 b^{2} {\color{red}{\int{\frac{\sin^{2}{\left(2 x \right)}}{4} d x}}}}{a^{2}} = \frac{2 b^{2} {\color{red}{\left(\frac{\int{\sin^{2}{\left(2 x \right)} d x}}{4}\right)}}}{a^{2}}$$

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

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

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 $$$f{\left(x \right)} = 1 - \cos{\left(4 x \right)}$$$:

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=1$$$:

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

Let $$$u=4 x$$$.

Then $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{4}$$$.

The integral can be rewritten as

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{4}$$$ and $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

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

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

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

Recall that $$$u=4 x$$$:

$$\frac{b^{2} \left(x - \frac{\sin{\left({\color{red}{u}} \right)}}{4}\right)}{4 a^{2}} = \frac{b^{2} \left(x - \frac{\sin{\left({\color{red}{\left(4 x\right)}} \right)}}{4}\right)}{4 a^{2}}$$

Therefore,

$$\int{\frac{b^{2} \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{a^{2}} d x} = \frac{b^{2} \left(x - \frac{\sin{\left(4 x \right)}}{4}\right)}{4 a^{2}}$$

Simplify:

$$\int{\frac{b^{2} \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{a^{2}} d x} = \frac{b^{2} \left(4 x - \sin{\left(4 x \right)}\right)}{16 a^{2}}$$

Add the constant of integration:

$$\int{\frac{b^{2} \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{a^{2}} d x} = \frac{b^{2} \left(4 x - \sin{\left(4 x \right)}\right)}{16 a^{2}}+C$$

$$$\int \frac{b^{2} \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{a^{2}}\, dx = \frac{b^{2} \left(4 x - \sin{\left(4 x \right)}\right)}{16 a^{2}} + C$$$A