# Homework Help: Integral help

1. Mar 25, 2005

### jaidon

this should be easy, but i'm struggling. i'm trying to help a friend solve:

integral(5x^3(1+x^2)^1/2), sorry i don't know how to do the integral sign.

i did a u sub with u=1+x^2, du=2xdx
and u-1=x^2

so:

(5/2)integral((u-1)(u)^1/2)

i get (1+x^2)^5/2-10/6(1+x^2)^3/2 +C

the answer says it should be 15/8(1+x^2)^4/3+C

i'm not sure what i have done wrong.

2. Mar 25, 2005

### whozum

********THIS WAS WRONG*********

Last edited: Mar 25, 2005
3. Mar 25, 2005

### whozum

Umm. hmph. Thats weird.

4. Mar 25, 2005

### jaidon

wow. that is unlike either answer i posted. i guess i should have stated that the only special way they have learned to solve integrals is with a u sub. did you see what exactly i did wrong in the way i tried to solve it? sometimes one needs an extra pair of eyes to find the error.

5. Mar 25, 2005

### Gamma

Wait a minute.

$$t = tan(u)$$

$$dt = tan(u)sec(u)du$$

is incorrect. dt = sec ^2 (u) du

6. Mar 25, 2005

### whozum

Oh I see where I messed up.

$$\int{5x^3\sqrt{1+x^2}}{dx}$$

You can do this with the trig substitution $$x = tan(u)$$
Then $$dx = sec^2(u)du$$

The integral becomes:

$$5\int{tan^3(u)\sqrt{1+tan^2(u)}sec^2(u)}{du}$$

$$1 + tan^2(u) = sec^2(u)$$

$$5\int{tan^3(u)\sqrt{sec^2(u)}sec^2(u)}{du}$$

$$5\int{tan^3(u)sec^3(u){du}$$

$$5\int{tan^2(u)sec^2(u)sec(u)tan(u)}{du}$$

$$5\int{(1-sec^2(u))sec^2(u)sec(u)tan(u)}{du}$$

Substituting $$t = sec(u), dt=sec(u)tan(u)du$$

$$5\int{(1-t^2)t^2dt = 5\int{t^2-t^4}{dt}$$

$$\frac{5t^3}{3}-\frac{5t^5}{5}$$

I butchered that first one but I don't see anything wrong with this one.

Last edited: Mar 25, 2005
7. Mar 25, 2005

### whozum

$$\frac{5}{2}\int(u-1)\sqrt{u}{du}$$

8. Mar 25, 2005

### jaidon

sorry, but i'm not sure how your answer is equivalent to the answer she was given ie)15/8(1+x^2)^4/3 +C. also, i guess i wasn't clear in my last post. they haven't done trig subs, by parts, partial fractions, etc, nothing like that yet. all they have done is u subs. this isn't even my question, but it is irritating me beyond belief. what is wrong with what i did?

9. Mar 25, 2005

### whozum

$$\frac{5}{2}\int(u-1)\sqrt{u}{du} = \frac{5}{2}\int{u^{3/2}-u^{1/2}}{du}$$

$$\frac{5}{2}( \frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2}) = u^{5/2}-\frac{5u^{3/2}}{3}$$

10. Mar 25, 2005

### Gamma

Whozum,

did not mean to find little mistakes. But I did this one and got the same results as Jidon. You have made couple of mistakes in the final two steps.

tan^u = sec^2 u -1 = t^2 - 1

$$5\int{(t^2 -1)t^2dt = 5\int{t^4-t^2}{dt}$$

$$t^5-\frac{5t^3}{3}$$

Jaidon: Your way is much simpler than trig sub. I don't see any thing wrong. Your book answer seems to be way off from what we are getting. Are you sure about that the given answer?

Last edited: Mar 25, 2005
11. Mar 25, 2005

### whozum

I dont see anything wrong with what you did, but the thing is when I derive their answer in maple I dont get anything near the original integral.

12. Mar 25, 2005

### whozum

Its ok gamma, better you find them than we let them run.

13. Mar 25, 2005

### jaidon

does anyone agree with me that maybe the answer given by my friend's sheet is incorrect?

14. Mar 25, 2005

### jaidon

15. Mar 25, 2005

### whozum

You said the answer was 15/8(1+x^2)^4/3

$$\frac{15}{8} (1+x^2)^{4/3}$$

Deriving that gives $$\frac{15}{8}*\frac{4}{3}(1+x^2)^{1/3}*2x$$ by the chain rule.

$$5x(1+x^2)^{1/3}$$

16. Mar 26, 2005

### jaidon

just wanted to let anyone who helped out with this know that my friend and i figured out, by looking at the derivative of the answer the prof gave, that there is a typo in the original question. it should not be 5x^3 (1+x^2)^1/2, but 5x (1+x^2)^1/3. it was written in such a way that i don't know anyone who would have read it as it should be. thought that was quite funny and just wanted to let my helpers know. thanks again for the help

jaidon

17. Mar 26, 2005

### whozum

In that case I integrated it right :)