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Math Help - trig substitutions for integrals

  1. #1
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    trig substitutions for integrals

    i have a test coming up and i wanted to double-check my answers:

    1. integral (b = 1, a = 0) x(sqrt (x^2 + 4)) dx
    i got 1/8 - 1/4sqrt(5) + c

    2. integral (du)/(u(sqrt(5-u^2)))
    1/5 ln u/sqrt(5) ln (sqrt(5) + u)/sqrt(5-^2)) + c


    3 integral (dt)/(sqrt(t^2 -6t +13))
    1/3 arctan (x-3)/2 + c

    if someone could tell me if i got these right...it would help me so muchh..thanks
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  2. #2
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    Hello !
    Quote Originally Posted by skabani View Post
    i have a test coming up and i wanted to double-check my answers:

    1. integral (b = 1, a = 0) x(sqrt (x^2 + 4)) dx
    i got 1/8 - 1/4sqrt(5) + c
    Nope ! There is a problem here !

    Substitute y=x^2+4

    after some working, we have :

    =\frac 12 \int_4^5 y^{1/2} ~dx (remember, \sqrt{a}=a^{1/2}

    But the primitive of y^n is \frac{y^{n+1}}{n+1}. Here, n=1/2 ---> n+1=3/2 ---> 1/(n+1)=2/3

    2. integral (du)/(u(sqrt(5-u^2)))
    1/5 ln u/sqrt(5) ln (sqrt(5) + u)/sqrt(5-^2)) + c
    I can't read your answer


    3 integral (dt)/(sqrt(t^2 -6t +13))
    1/3 arctan (x-3)/2 + c
    It's 1/2 ...

    To check your answers, use this : Wolfram Mathematica Online Integrator
    or any mathematical software or calculator !
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  3. #3
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    okay,

    so is number one: 111/5

    two, i re-did it and got: (1/sqrt(5)) ln|csc x - cot x|+ c
    i know we have to substitute the original variables back into the answer to get it in terms of u, but im not sure how to do that..

    three: im not sure how to get 1/2, do you mind showing me

    i tried the website, but all it does is factors out the variable and gives me the original integral...thanks anyways though
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  4. #4
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    For the second problem, you should not be getting any natural logs in the answer.

    The integral is:
    \int \frac{u}{\sqrt{5 - u^2}} ~ du

    By multiplying by -1 inside and -1 outside you get

    -\int \frac{-u}{\sqrt{5 - u^2}} ~ du

    and the integral evaluates to
    -\sqrt{5 - u^2} + C.
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  5. #5
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    Quote Originally Posted by icemanfan View Post
    For the second problem, you should not be getting any natural logs in the answer.

    The integral is:
    \int \frac{u}{\sqrt{5 - u^2}} ~ du

    By multiplying by -1 inside and -1 outside you get

    -\int \frac{-u}{\sqrt{5 - u^2}} ~ du

    and the integral evaluates to
    -\sqrt{5 - u^2} + C.
    i thought we were supposed to use trig substitution- set u=sqrt(5) sin x
    and solve from there?
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  6. #6
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    Quote Originally Posted by skabani View Post
    i thought we were supposed to use trig substitution- set u=sqrt(5) sin x
    and solve from there?
    You can use trig substitution sometimes when you have \frac{1}{\sqrt{a^2 - x^2}} or similar situations, but in this case, the integral is in the y^n ~ dy form, where y = 5 - u^2, n = -\frac{1}{2}, dy = -2u ~ du. It is easier to use the y^n ~ dy form if you can.
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  7. #7
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    Quote Originally Posted by icemanfan View Post
    You can use trig substitution sometimes when you have \frac{1}{\sqrt{a^2 - x^2}} or similar situations, but in this case, the integral is in the y^n ~ dy form, where y = 5 - u^2, n = -\frac{1}{2}, dy = -2u ~ du. It is easier to use the y^n ~ dy form if you can.
    oh, okay, and do you mind checking number 1 for me..i got 111/5

    thanks so much!!
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  8. #8
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    I thought the second integral was:
    \int \frac{du}{u\sqrt{5-u^2}}

    From what skabani wrote in his first post, it looks that way. Can you clear this up Skabani?
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  9. #9
    Super Member 11rdc11's Avatar
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    I'm confused is the second integral

    1st form \int \frac{du}{u\sqrt{5-u^2}}


    2nd form \int \frac{u~du}{\sqrt{5-u^2}}
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  10. #10
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    Quote Originally Posted by Chop Suey View Post
    I thought the second integral was:
    \int \frac{du}{u\sqrt{5-u^2}}

    From what skabani wrote in his first post, it looks that way. Can you clear this up Skabani?
    Oops, I think you're right. I read the integral wrong. Looks like trig substitution is the way to go.
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  11. #11
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    Quote Originally Posted by Chop Suey View Post
    I thought the second integral was:
    \int \frac{du}{u\sqrt{5-u^2}}

    From what skabani wrote in his first post, it looks that way. Can you clear this up Skabani?

    actually, you are right, it is what chop suey said...
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  12. #12
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    Then substitute t = \frac{1}{\sqrt{5-u^2}} and see where that gets you.
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  13. #13
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    O ok well then you can approach it like this

    divide top and bottom by \sqrt{5}

    which leaves

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

    and then substitute

    sin\theta = \frac{u}{\sqrt{5}}

    du= \sqrt{5}\cos\theta

    u = \sqrt{5}\sin\theta

    and use the identity

    1- \sin^2\theta = \cos^2\theta

    \frac{1}{\sqrt{5}} \int \frac{\sqrt{5}\cos\theta}{(\sqrt{5}\sin\theta)(\co  s\theta)}~d\theta = \frac{1}{\sqrt{5}} \int \csc\theta d\theta = \frac{-\ln(\csc\theta + \cot\theta)}{\sqrt{5}}

    Now just back sub
    Last edited by 11rdc11; September 23rd 2008 at 03:04 PM.
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  14. #14
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    Quote Originally Posted by Chop Suey View Post
    Then substitute t = \frac{1}{\sqrt{5-u^2}} and see where that gets you.
    but doing so would leave me with 1/ut right?

    hm...im not sure....can you help me out..
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  15. #15
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    Quote Originally Posted by 11rdc11 View Post
    O ok well then you can approach it like this

    divide top and bottom by \sqrt{5}

    which leaves

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

    and then substitute

    sin\theta = \frac{u}{\sqrt{5}}

    du= \sqrt{5}\cos\theta

    u = \sqrt{5}\sin\theta

    and use the identity

    1- \sin^2\theta = \cos^2\theta

    \frac{1}{\sqrt{5}} \int \frac{\sqrt{5}\cos\theta}{(\sqrt{5}\sin\theta)(\co  s\theta)}~d\theta = \frac{1}{\sqrt{5}} \int \csc\theta d\theta = \frac{-\ln(\csc\theta + \cot\theta)}{\sqrt{5}}

    Now just back sub
    okay..that makes sense..thankss!
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