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

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

Let $$$u=x^{2}$$$.

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

Thus,

$${\color{red}{\int{x^{3} \sin{\left(x^{2} \right)} d x}}} = {\color{red}{\int{\frac{u \sin{\left(u \right)}}{2} d u}}}$$

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

$${\color{red}{\int{\frac{u \sin{\left(u \right)}}{2} d u}}} = {\color{red}{\left(\frac{\int{u \sin{\left(u \right)} d u}}{2}\right)}}$$

For the integral $$$\int{u \sin{\left(u \right)} d u}$$$, use integration by parts $$$\int \operatorname{t} \operatorname{dv} = \operatorname{t}\operatorname{v} - \int \operatorname{v} \operatorname{dt}$$$.

Let $$$\operatorname{t}=u$$$ and $$$\operatorname{dv}=\sin{\left(u \right)} du$$$.

Then $$$\operatorname{dt}=\left(u\right)^{\prime }du=1 du$$$ (steps can be seen ») and $$$\operatorname{v}=\int{\sin{\left(u \right)} d u}=- \cos{\left(u \right)}$$$ (steps can be seen »).

Thus,

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

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

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

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

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

Recall that $$$u=x^{2}$$$:

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

Therefore,

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

Add the constant of integration:

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

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