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Thread: Integrals using U-Substitution

  1. #1
    Member FalconPUNCH!'s Avatar
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    Integrals using U-Substitution

    Can anyone check if I'm doing these correct

    1. $\displaystyle \int{\frac{(lnx)^2}{x}dx}$

    $\displaystyle U = lnx , DU = \frac{dx}{x}$

    $\displaystyle \int{u}{du}$

    $\displaystyle = \frac{1}{2}(u^2)$

    $\displaystyle = \frac {1}{2}(lnx)^2$

    2. $\displaystyle \int\frac{x}{(x^2+2)^2}dx$

    $\displaystyle U = x^2 + 1 , \frac{1}{2}DU = xdx$

    $\displaystyle = \frac{1}{1}\int\frac{1}{u^2}du$

    $\displaystyle = \frac{1}{2}ln(u^2) $

    $\displaystyle = \frac{1}{2}ln|(x^2)+1|$

    I know I made mistakes I just don't know where.
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  2. #2
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    You made a mistake in #1:

    after you substitute u = ln(x) you should get:

    $\displaystyle \int u^2 du$

    And in #2 you should make $\displaystyle u = x^2 + 2$.
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  3. #3
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    Quote Originally Posted by TrevorP View Post
    You made a mistake in #1:

    after you substitute u = ln(x) you should get:

    $\displaystyle \int u^2 du$

    And in #2 you should make $\displaystyle u = x^2 + 2$.
    Yes, and in the second one, you would then be integrating $\displaystyle u^{-2}$ which is $\displaystyle -u^{-1}$
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    Quote Originally Posted by FalconPUNCH! View Post
    $\displaystyle = \frac{1}{1}\int\frac{1}{u^2}du$

    $\displaystyle = \frac{1}{2}ln(u^2) $
    This is a common mistake.

    It is $\displaystyle \int {\frac{1}
    {u}\,du} = \ln \left| u \right| + k.$ For $\displaystyle \int {\frac{1}
    {{u^2 }}\,du}$ write the integrand as $\displaystyle u^{-2}$ and apply the power rule.
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  5. #5
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    Quote Originally Posted by FalconPUNCH! View Post
    Can anyone check if I'm doing these correct

    1. $\displaystyle \int{\frac{(lnx)^2}{x}dx}$

    $\displaystyle U = lnx , DU = \frac{dx}{x}$

    $\displaystyle \int{u}{du}$

    $\displaystyle = \frac{1}{2}(u^2)$

    $\displaystyle = \frac {1}{2}(lnx)^2$

    2. $\displaystyle \int\frac{x}{(x^2+2)^2}dx$

    $\displaystyle U = x^2 + 1 , \frac{1}{2}DU = xdx$

    $\displaystyle = \frac{1}{1}\int\frac{1}{u^2}du$

    $\displaystyle = \frac{1}{2}ln(u^2) $

    $\displaystyle = \frac{1}{2}ln|(x^2)+1|$

    I know I made mistakes I just don't know where.
    Most of what needs to be said has been said and said well.

    My 2 cents is that if the original questions asked for the integral, the arbitrary constant of integration must be included in the answer.
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  6. #6
    Member FalconPUNCH!'s Avatar
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    Thanks guys. Sorry I didn't respond back until today. Well I have more problems which are harder and I don't really know where to start with some of them.

    1. $\displaystyle \int\frac{x^2}{\sqrt{1-x}}dx$

    For this one I don't know here to begina

    2. $\displaystyle \frac{sinx}{1+cos^{2} x} dx$

    3. $\displaystyle \frac{sin2x}{1+cos^{2}x}dx$

    For these two I was never that good in trig. so it's hard for me to understand where to start.

    4.$\displaystyle \int_{0}^{1} (x)\sqrt{x-1}dx$

    I'm not really looking for answers but I am looking for how to start these problems. Thank you guys
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    Use u-substitution for the first one:

    $\displaystyle u=1-x$
    $\displaystyle du=-dx$

    $\displaystyle x=1-u$
    $\displaystyle x^2=(1-u)^2 \Rightarrow u^2 - 2u + 1$

    Your integrand then becomes: $\displaystyle \frac{-u^2+2u-1}{\sqrt{u}}$

    Separate each term into its own fraction and apply the power rules for integration to each.
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  8. #8
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    Quote Originally Posted by FalconPUNCH! View Post

    1. $\displaystyle \int\frac{x^2}{\sqrt{1-x}}dx$
    Substitute $\displaystyle u^2=1-x.$


    Quote Originally Posted by FalconPUNCH! View Post
    2. $\displaystyle \frac{sinx}{1+cos^{2} x} dx$
    This is an arctangent. First substitute $\displaystyle u=\cos x.$

    Quote Originally Posted by FalconPUNCH! View Post
    3. $\displaystyle \frac{sin2x}{1+cos^{2}x}dx$
    This was posted before, you just need to note that

    $\displaystyle \left(1+ {\cos ^2 x} \right)' = - 2\sin x\cos x = - \sin 2x.$

    Quote Originally Posted by FalconPUNCH! View Post
    4.$\displaystyle \int_{0}^{1} (x)\sqrt{x-1}dx$
    Substitute $\displaystyle u^2=x-1.$ (Do you know Barrow's Rule? When makin' a change of variables you need to get the new integration limits for $\displaystyle u.$)
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  9. #9
    Member FalconPUNCH!'s Avatar
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    Thank you all for your help. I will try to solve them and come back tomorrow or after tomorrow with my results.

    Edit: I do not know Barrow's Rule.
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  10. #10
    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by FalconPUNCH! View Post


    3. $\displaystyle \frac{sin2x}{1+cos^{2}x}dx$

    For this one is u = sin2x?
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    Quote Originally Posted by FalconPUNCH! View Post
    For this one is u = sin2x?
    No, I would say not. There is a trig identity that says sin2x = 2sinxcosx.

    If you fill that in then set u = cos x, you should get somewhere.
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  12. #12
    Member FalconPUNCH!'s Avatar
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    Ok I sort of understood. I'm not to good with trig but here's what I did

    $\displaystyle \int\frac{2sinxcosx}{1+cos^{2}x}$

    $\displaystyle u = cos x ; du = -sinxdx$

    $\displaystyle -2\int\frac{u}{1+u^2}du$

    $\displaystyle -2tan^{-1}(cos^{2}x) + c $
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  13. #13
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    Yes the only problem I see with that is that you should have -2Arctan(Arccos(x))
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  14. #14
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    Quote Originally Posted by TrevorP View Post
    Yes the only problem I see with that is that you should have -2Arctan(Arccos(x))

    Sorry, I think that's wrong.

    You had $\displaystyle \int \frac{2sinxcosx}{1+cos^2x}dx$

    Integrating this gives $\displaystyle -ln(1+cos^2x)+C$

    Your answer does not seem to be equivalent.
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  15. #15
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    Quote Originally Posted by a tutor View Post
    Sorry, I think that's wrong.

    You had $\displaystyle \int \frac{2sinxcosx}{1+cos^2x}dx$

    Integrating this gives $\displaystyle -ln(1+cos^2x)+C$

    Your answer does not seem to be equivalent.

    Ohh, you're right.
    Last edited by mr fantastic; Jul 10th 2011 at 04:31 PM.
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