Integral help

  • Thread starter jaidon
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  • #1
42
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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.
 

Answers and Replies

  • #2
2,210
1
********THIS WAS WRONG*********
 
Last edited:
  • #3
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Umm. hmph. Thats weird.
 
  • #4
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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
357
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Wait a minute.

[tex] t = tan(u) [/tex]

[tex] dt = tan(u)sec(u)du [/tex]

is incorrect. dt = sec ^2 (u) du
 
  • #6
2,210
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Oh I see where I messed up.

[tex] \int{5x^3\sqrt{1+x^2}}{dx} [/tex]

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

The integral becomes:

[tex] 5\int{tan^3(u)\sqrt{1+tan^2(u)}sec^2(u)}{du} [/tex]

[tex] 1 + tan^2(u) = sec^2(u) [/tex]

[tex] 5\int{tan^3(u)\sqrt{sec^2(u)}sec^2(u)}{du} [/tex]

[tex] 5\int{tan^3(u)sec^3(u){du} [/tex]

[tex] 5\int{tan^2(u)sec^2(u)sec(u)tan(u)}{du} [/tex]

[tex] 5\int{(1-sec^2(u))sec^2(u)sec(u)tan(u)}{du}[/tex]

Substituting [tex] t = sec(u), dt=sec(u)tan(u)du [/tex]

[tex] 5\int{(1-t^2)t^2dt = 5\int{t^2-t^4}{dt}[/tex]

[tex] \frac{5t^3}{3}-\frac{5t^5}{5} [/tex]

I butchered that first one but I don't see anything wrong with this one.
 
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  • #7
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[tex] \frac{5}{2}\int(u-1)\sqrt{u}{du} [/tex]
 
  • #8
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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
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[tex] \frac{5}{2}\int(u-1)\sqrt{u}{du} = \frac{5}{2}\int{u^{3/2}-u^{1/2}}{du}[/tex]

[tex] \frac{5}{2}( \frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2}) = u^{5/2}-\frac{5u^{3/2}}{3}[/tex]
 
  • #10
357
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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

[tex] 5\int{(t^2 -1)t^2dt = 5\int{t^4-t^2}{dt}[/tex]



[tex] t^5-\frac{5t^3}{3} [/tex]


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?
 
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  • #11
2,210
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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
2,210
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Its ok gamma, better you find them than we let them run.
 
  • #13
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does anyone agree with me that maybe the answer given by my friend's sheet is incorrect?
 
  • #14
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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
 
  • #15
2,210
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You said the answer was 15/8(1+x^2)^4/3

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

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

[tex] 5x(1+x^2)^{1/3} [/tex]
 
  • #16
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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
2,210
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In that case I integrated it right :)
 

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