Take the 2-minute tour ×
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.

$\displaystyle\int{t^2}{{\sqrt[3]{t^3-1}}}\,dt$. Can someone give me a hint on how to solve this problem?

share|improve this question

closed as off-topic by Mike Miller, Sanath Devalapurkar, Hans Engler, Michael Albanese, user61527 Jun 10 '14 at 2:47

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Mike Miller, Sanath Devalapurkar, Hans Engler, Michael Albanese, Community
If this question can be reworded to fit the rules in the help center, please edit the question.

Substitution. Let $u=t^3-1$. –  André Nicolas Jun 10 '14 at 0:09

3 Answers 3

up vote 3 down vote accepted

Substitute the following

  • $u=t^3-1$
  • $du=3t^2 \, dt \implies \frac{du}{3}= t^2 \, dt$

So that we may now perform the integration by $u$-substitution. \begin{align}\int t^2 \sqrt[3]{t^3-1} \, dt &= \int \sqrt[3]{t^3-1} \, \underbrace{t^2 \, dt}_{du/3}\\ &= \int \sqrt[3] u \, \frac{du}{3} \\ &=\frac 13 \int u^{\frac 13} \, du \\ &=\frac 13 \frac 34u^{\frac 43} +C \\ &=\frac 14(t^3-1)^{\frac 43} +C\end{align}

share|improve this answer
Do not provide the whole answer. Allow the OP to do some work. –  Sanath Devalapurkar Jun 10 '14 at 0:19
This isn't a homework problem, or at least it doesn't have the (homework) tag on it. So I am free to provide the whole answer. –  dragon Jun 10 '14 at 0:19
Yes, but simple problems such as these usually are homework. –  Sanath Devalapurkar Jun 10 '14 at 0:20
I'm a little confused on the $3/4$ part –  user124557 Jun 10 '14 at 0:30
Anti-power rule: $$\int y^{\frac 13} \, du= \frac{y^{\frac 13 + 1}}{\frac 13 + 1} + C = \frac{y^{\frac 43}}{\frac 43} + C = \frac 34 u^{\frac 43} + C $$ –  dragon Jun 10 '14 at 0:31

The usual substitution is $u=t^3-1$. Alternately, let $u^3=t^3-1$. Then $3u^2\,du=3t^2\,dt$, so we want $\int u^3\,du$.

share|improve this answer

Let $u=t^3-1\implies du=3t^2dt$. Hence, $$\dfrac{1}{3}\int \sqrt[3]{t^3-1}(3t^2dt)=\dfrac{1}{3}\int \sqrt[3]{u}du$$ You should be able to do this.

share|improve this answer

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