Integral of $$$x^{3} \cos{\left(x^{2} \right)}$$$
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Find $$$\int x^{3} \cos{\left(x^{2} \right)}\, dx$$$.
Solution
Let $$$u=x^{2}$$$.
Then $$$du=\left(x^{2}\right)^{\prime }dx = 2 x dx$$$ (steps can be seen »), and we have that $$$x dx = \frac{du}{2}$$$.
So,
$${\color{red}{\int{x^{3} \cos{\left(x^{2} \right)} d x}}} = {\color{red}{\int{\frac{u \cos{\left(u \right)}}{2} d u}}}$$
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 $$$f{\left(u \right)} = u \cos{\left(u \right)}$$$:
$${\color{red}{\int{\frac{u \cos{\left(u \right)}}{2} d u}}} = {\color{red}{\left(\frac{\int{u \cos{\left(u \right)} d u}}{2}\right)}}$$
For the integral $$$\int{u \cos{\left(u \right)} d u}$$$, use integration by parts $$$\int \operatorname{c} \operatorname{dv} = \operatorname{c}\operatorname{v} - \int \operatorname{v} \operatorname{dc}$$$.
Let $$$\operatorname{c}=u$$$ and $$$\operatorname{dv}=\cos{\left(u \right)} du$$$.
Then $$$\operatorname{dc}=\left(u\right)^{\prime }du=1 du$$$ (steps can be seen ») and $$$\operatorname{v}=\int{\cos{\left(u \right)} d u}=\sin{\left(u \right)}$$$ (steps can be seen »).
The integral can be rewritten as
$$\frac{{\color{red}{\int{u \cos{\left(u \right)} d u}}}}{2}=\frac{{\color{red}{\left(u \cdot \sin{\left(u \right)}-\int{\sin{\left(u \right)} \cdot 1 d u}\right)}}}{2}=\frac{{\color{red}{\left(u \sin{\left(u \right)} - \int{\sin{\left(u \right)} d u}\right)}}}{2}$$
The integral of the sine is $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:
$$\frac{u \sin{\left(u \right)}}{2} - \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{2} = \frac{u \sin{\left(u \right)}}{2} - \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{2}$$
Recall that $$$u=x^{2}$$$:
$$\frac{\cos{\left({\color{red}{u}} \right)}}{2} + \frac{{\color{red}{u}} \sin{\left({\color{red}{u}} \right)}}{2} = \frac{\cos{\left({\color{red}{x^{2}}} \right)}}{2} + \frac{{\color{red}{x^{2}}} \sin{\left({\color{red}{x^{2}}} \right)}}{2}$$
Therefore,
$$\int{x^{3} \cos{\left(x^{2} \right)} d x} = \frac{x^{2} \sin{\left(x^{2} \right)}}{2} + \frac{\cos{\left(x^{2} \right)}}{2}$$
Simplify:
$$\int{x^{3} \cos{\left(x^{2} \right)} d x} = \frac{x^{2} \sin{\left(x^{2} \right)} + \cos{\left(x^{2} \right)}}{2}$$
Add the constant of integration:
$$\int{x^{3} \cos{\left(x^{2} \right)} d x} = \frac{x^{2} \sin{\left(x^{2} \right)} + \cos{\left(x^{2} \right)}}{2}+C$$
Answer
$$$\int x^{3} \cos{\left(x^{2} \right)}\, dx = \frac{x^{2} \sin{\left(x^{2} \right)} + \cos{\left(x^{2} \right)}}{2} + C$$$A