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

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

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

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

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

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

Expand the expression:

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

Integrate term by term:

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

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

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

For the integral $$$\int{x \cos{\left(2 x \right)} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=x$$$ and $$$\operatorname{dv}=\cos{\left(2 x \right)} dx$$$.

Then $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{\cos{\left(2 x \right)} d x}=\frac{\sin{\left(2 x \right)}}{2}$$$ (steps can be seen »).

So,

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

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

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

Let $$$u=2 x$$$.

Then $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{2}$$$.

Thus,

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

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

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

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

Recall that $$$u=2 x$$$:

$$\frac{x^{2}}{4} - \frac{x \sin{\left(2 x \right)}}{4} - \frac{\cos{\left({\color{red}{u}} \right)}}{8} = \frac{x^{2}}{4} - \frac{x \sin{\left(2 x \right)}}{4} - \frac{\cos{\left({\color{red}{\left(2 x\right)}} \right)}}{8}$$

Therefore,

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

Add the constant of integration:

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

$$$\int x \sin^{2}{\left(x \right)}\, dx = \left(\frac{x^{2}}{4} - \frac{x \sin{\left(2 x \right)}}{4} - \frac{\cos{\left(2 x \right)}}{8}\right) + C$$$A