1. ## Integral Help

$\displaystyle \int5x\sqrt[3]{1-x^2}dx$

I am required to use U subsitiution.

u= $\displaystyle 1-x^2$
du= $\displaystyle -2x$ $\displaystyle dx$

So the problem became....

$\displaystyle \frac{-2}{5} \int u^{1/3} du$

$\displaystyle \frac{-3}{10} u^{4/3} + C$

This seems correct to me, however the answer my book gives is...
$\displaystyle \frac{-15}{8} u^{4/3} + C$

Of course I will replace the u in the final answer, but I cannot figure out how the book did this problem. Can someone please tell me what I did wrong.

Thanks so much!

2. Originally Posted by liz155
$\displaystyle \int5x\sqrt[3]{1-x^2}dx$

I am required to use U subsitiution.

u= $\displaystyle 1-x^2$
du= $\displaystyle -2x$ $\displaystyle dx$

So the problem became....

$\displaystyle \frac{-2}{5} \int u^{1/3} du$

$\displaystyle \frac{-3}{10} u^{4/3} + C$

This seems correct to me, however the answer my book gives is...
$\displaystyle \frac{-15}{8} u^{4/3} + C$

Of course I will replace the u in the final answer, but I cannot figure out how the book did this problem. Can someone please tell me what I did wrong.

Thanks so much!
the constant you should have in front of the new integral is $\displaystyle - \frac 52$, try again

3. How did you get -5/2?

I thought that in order to replace the 5x dx with the du of -2x dx, I would have to multiply by some constant. I thought I had to get the 5x dx part equal to the du, but have a constant out in front. I multiplied by -2/5 so the 5's cancelled out and only -2 was left. I'm confused on how you get -5/2...

Then solving the rest of the problem and reducing fractions led to the -3/10.

4. Originally Posted by liz155
How did you get -5/2?

I thought that in order to replace the 5x dx with the du of -2x dx, I would have to multiply by some constant. I thought I had to get the 5x dx part equal to the du, but have a constant out in front. I multiplied by -2/5 so the 5's cancelled out and only -2 was left. I'm confused on how you get -5/2...

Then solving the rest of the problem and reducing fractions led to the -3/10.

$\displaystyle du = -2x~dx$

$\displaystyle \Rightarrow - \frac 12 du = x~dx$

so our integral becomes:

$\displaystyle 5 \cdot \left( - \frac 12 \right) \int u^{1/3}~du$

and continue...

you should not seek to cancel the 5, it was a constant that was there, leave it alone. you should seek to cancel the -2 that you got from the substitution, so you multiply by -1/2

5. Thanks soooo much!!! I see what I did wrong and after finishing the problem, I got the right answer!

Thank you!!