Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In a solutions manual I see they integrate this $\frac{1}{2}r(4-r^2)^2$ and in the next this is $-\frac{1}{12}(4-r^2)^3$. Is this possible without working out the parentheses?

Can someone explain this step? Thanks!

share|cite|improve this question
Let $u=4-r^2$. Or else guess the answer will look like $k(4-r^2)^3$, and differentiate to see what $k$ should be. – André Nicolas Apr 12 '14 at 20:00

1 Answer 1

Yes, definitely. We are given

$$\int \dfrac{1}{2}r(4 - r^2)^2\,dr$$

Set $u = 4 - r^2$. Then, $du = -2r\,dr$, which gives

$$-\dfrac{1}{4} \int u^2\,du = -\dfrac{1}{12} u^3 + \mbox{C}$$

Thus, we have $-\frac{1}{12}(4 - r^2)^3 + \mbox{C}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.