Page 1 of 2 12 LastLast
Results 1 to 15 of 16

Thread: trig substitution

  1. #1
    Member
    Joined
    May 2010
    From
    WI - USA
    Posts
    129

    trig substitution

    Good afternoon,

    I have worked out the following problem and arrived at the following answer. Would someone mind verifying this for me?

    $\displaystyle
    {\int{\frac{\sqrt{4-x^2}}{x^2}}dx
    $


    let $\displaystyle x=2sin\theta$
    let $\displaystyle dx=2cos\theta d\theta$


    I arrive at


    $\displaystyle
    {\int{\frac{\sqrt{4-x^2}}{x^2}}
    =\frac{1}{2}\ln\mid\frac{2+x}{\sqrt{4-x^2}}\mid+C$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    7
    Awards
    2
    Hmm. I don't get your result. Could you please post a few more steps?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2010
    From
    WI - USA
    Posts
    129
    Sure,

    Using the substition...

    $\displaystyle
    x=2sin\theta
    $
    $\displaystyle
    dx=2cos\theta d\theta
    $

    I get...

    $\displaystyle
    \sqrt{4-x^2} = \sqrt{4-4sin^2\theta} = 2\sqrt{1-sin^2\theta} = 2\sqrt{cos^2\theta} = 2cos\theta
    $

    So...

    $\displaystyle
    \int{\frac{2cos\theta}{4cos^2\theta}d\theta}
    $

    Wait... Should that have been...

    $\displaystyle \int{\frac{d\theta}{4cos^2\theta}}$

    This must have been where I went wrong. What should this have been, and why?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    [QUOTE=MechEng;530685]Good afternoon,

    I have worked out the following problem and arrived at the following answer. Would someone mind verifying this for me?

    $\displaystyle
    {\int{\frac{\sqrt{4-x^2}}{x^2}}dx
    $


    let $\displaystyle x=2sin\theta$
    let $\displaystyle dx=2cos\theta d\theta$


    The substitution is good!!!

    ...but the final answer is wrong!

    (Take the derivative of what you got, is it the integrand?)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    7
    Awards
    2
    Double-check your denominator.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member slider142's Avatar
    Joined
    May 2009
    From
    Brooklyn, NY
    Posts
    72
    Quote Originally Posted by MechEng View Post
    ...
    $\displaystyle
    \int{\frac{2cos\theta}{4cos^2\theta}d\theta}
    $
    ...
    Your numerator is indeed $\displaystyle 2\cos \theta$ but your denominator is $\displaystyle x^2 = (2\sin \theta)^2$, and don't forget to replace dx with $\displaystyle 2\cos\theta d\theta$.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2
    Totally not right.

    Quote Originally Posted by MechEng View Post
    $\displaystyle
    {\int{\frac{\sqrt{4-x^2}}{x^2}}
    =\frac{1}{2}\ln\mid\frac{2+x}{\sqrt{4-x^2}}\mid+C$
    $\displaystyle \dfrac{1}{2}\ln|\dfrac{2+x}{\sqrt{4-x^2}}| = \dfrac{1}{2}\text{arctanh}\dfrac{x}{2}\Rightarrow \bigg\{\dfrac{1}{2}\ln|\dfrac{2+x}{\sqrt{4-x^2}}|\bigg\}' = \dfrac{1}{1-4x^2}.$
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1

    no,

    it's:

    $\displaystyle \int cot^2\theta d\theta = -\theta- cot\theta +C$
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2
    Quote Originally Posted by Also sprach Zarathustra View Post
    No
    What? The OP has not yet responded. I don't know what on Earth you are referring to.

    it's:

    $\displaystyle \int cot^2\theta d\theta = -\theta + cot\theta +C$
    The plus should be minus.

    EDIT. I see you edited your post.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    You are blind or what??? Where are you see minus??? hahaha...
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    May 2010
    From
    WI - USA
    Posts
    129
    Good afternoon All,

    I am sitting at work without my notes today, trying to pickup where I left off on this one.

    Using the "correct" substitution I get the following:

    $\displaystyle
    \int{\frac{2cos\theta}{(2sin\theta)^2}(2cos\theta) d\theta
    $

    Can this be reduced to:

    $\displaystyle
    \int{\frac{(2cos\theta)^2}{(2sin\theta)^2}d\theta
    $

    And then:

    $\displaystyle
    \int{cot^2\theta}d\theta
    $

    And then:

    $\displaystyle
    -\theta-cot^2\theta+c
    $

    Since $\displaystyle x=2sin\theta$ we have $\displaystyle cot\theta=\frac{\sqrt{4-x^2}}{x}$ and $\displaystyle \theta=sin^{-1}(\frac{x}{2})$

    So:

    $\displaystyle -sin^{-1}(\frac{x}{2})-(\frac{\sqrt{4-x^2}}{x})^2+C$

    And:

    $\displaystyle
    -sin^{-1}(\frac{x}{2})-\frac{\mid4-x^2\mid}{x^2}+C
    $


    Am I totally off target again?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2
    Quote Originally Posted by MechEng View Post
    $\displaystyle
    \int{cot^2\theta}d\theta
    $

    And then:

    $\displaystyle
    -\theta-cot^2\theta+c
    $
    Are you sure?
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Member
    Joined
    May 2010
    From
    WI - USA
    Posts
    129
    Should that look more like:

    $\displaystyle \int{cot^2\theta}d\theta$ = $\displaystyle \ln{\mid sin^2\theta\mid+C$

    I guess I'm confused by the $\displaystyle cot^2\theta$. Well, that and I am trying to do this without notes of any kind... between phone calls and e-mails.

    Thank you.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3702
    note the identity ...

    $\displaystyle \displaystyle \cot^2{t} = \csc^2{t} - 1$

    sub in for $\displaystyle \cot^2{t}$ in the integrand ...

    $\displaystyle \displaystyle \int \csc^2{t} - 1 \, dt = -\cot{t} - t + C$

    $\displaystyle \displaystyle t = \arcsin\left(\frac{x}{2}\right)$

    back substitute ...

    $\displaystyle \displaystyle -\frac{\sqrt{4-x^2}}{x} - \arcsin\left(\frac{x}{2}\right) + C$
    Last edited by skeeter; Jun 29th 2010 at 12:32 PM. Reason: fixed typo
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Member
    Joined
    May 2010
    From
    WI - USA
    Posts
    129
    Quote Originally Posted by skeeter View Post

    sub in for $\displaystyle \cos^2{t}$ in the integrand ...
    Where am I substituting this?

    Oh, should that have read:

    Quote Originally Posted by skeeter View Post

    sub in for $\displaystyle \cot^2{t}$ in the integrand ...
    I'm getting really hung up on my trig identities.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Trig Substitution
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Feb 23rd 2010, 01:12 PM
  2. trig substitution
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Feb 6th 2010, 01:44 PM
  3. Trig substitution
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 4th 2009, 12:37 PM
  4. Trig Substitution
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Mar 25th 2009, 06:11 AM
  5. Trig Substitution Help
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Mar 1st 2009, 11:29 PM

Search Tags


/mathhelpforum @mathhelpforum