Solving an Integral: Helping a Friend

[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.

 

[whozum]

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

 

[whozum]

Umm. hmph. Thats weird.

 

[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.

 

[Gamma]

Wait a minute.

t = tan(u)

dt = tan(u)sec(u)du

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

 

[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.

 

[whozum]

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

 

[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?

 

[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}

 

[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?

 

[whozum]

I don't see anything wrong with what you did, but the thing is when I derive their answer in maple I don't get anything near the original integral.

 

[whozum]

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

 

[jaidon]

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

 

[jaidon]

sorry Gamma, missed your post about the answer being wrong. glad to here that i am not way off base on this. thanks all for your input

 

[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}

 

[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

 

[whozum]

In that case I integrated it right :)