Integral of $$$x^{2} \sin{\left(x^{3} \right)}$$$
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Find $$$\int x^{2} \sin{\left(x^{3} \right)}\, dx$$$.
Solution
Let $$$u=x^{3}$$$.
Then $$$du=\left(x^{3}\right)^{\prime }dx = 3 x^{2} dx$$$ (steps can be seen »), and we have that $$$x^{2} dx = \frac{du}{3}$$$.
Therefore,
$${\color{red}{\int{x^{2} \sin{\left(x^{3} \right)} d x}}} = {\color{red}{\int{\frac{\sin{\left(u \right)}}{3} d u}}}$$
Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{3}$$$ and $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:
$${\color{red}{\int{\frac{\sin{\left(u \right)}}{3} d u}}} = {\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{3}\right)}}$$
The integral of the sine is $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:
$$\frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{3} = \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{3}$$
Recall that $$$u=x^{3}$$$:
$$- \frac{\cos{\left({\color{red}{u}} \right)}}{3} = - \frac{\cos{\left({\color{red}{x^{3}}} \right)}}{3}$$
Therefore,
$$\int{x^{2} \sin{\left(x^{3} \right)} d x} = - \frac{\cos{\left(x^{3} \right)}}{3}$$
Add the constant of integration:
$$\int{x^{2} \sin{\left(x^{3} \right)} d x} = - \frac{\cos{\left(x^{3} \right)}}{3}+C$$
Answer
$$$\int x^{2} \sin{\left(x^{3} \right)}\, dx = - \frac{\cos{\left(x^{3} \right)}}{3} + C$$$A