TITLE: Evaluate $\int x^2 \sin(7x^3)dx$
QUESTION [7 upvotes]: Evaluate $$\int x^2 \sin(7x^3)dx.$$

Can somebody check my solution? Thanks!
Let $u = 7x^3$, then $du = 21x^2dx$ and so $\dfrac{du}{21x^2}=dx$.
Thus we have: $$\int x^2 \sin(7x^3)dx=\frac{1}{21} \int \sin(u)du = \frac{-1}{21}\cos(u)+C = \frac{-1}{21}\cos(7x^3)+C$$

REPLY [6 votes]: Your solution is perfect! It could not be done better.
Although, you could check it by your self by taking the derivative of your result.