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Math Help - Need help with a u-substitution problem

  1. #1
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    Need help with a u-substitution problem

    integrate xsecx^2tanx^2

    (notice: not, xsec^2xtan^x)
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  2. #2
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    Put z=x^2, and the integral you'll get is a basic one, it comes from the derivative of secant.
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  3. #3
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    i came out to 2secx+c. Is that right?
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  4. #4
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    Quote Originally Posted by plooms View Post
    i came out to 2secx+c. Is that right?
    No, you have to substitute back. And the 2 is in the denominator !

    Because when subbing y=x, dy=2x dx ---> dx=dy/(2x)

    a way to check a result : Wolfram Mathematica Online Integrator
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  5. #5
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    got it! and thanks for the website
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  6. #6
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    Integrate 2/[x^(1/4)+x]

    How would i go about finding a substitution for that?
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  7. #7
    Moo
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    Quote Originally Posted by plooms View Post
    Integrate 2/[x^(1/4)+x]

    How would i go about finding a substitution for that?
    Here is what I would do :
    Substituting y=x^{1/4}+x ? Not good. The dy will have a nasty expression !

    Substituting y=x^{1/4} ? Not bad... But it would be something weird though

    Another method ? Why not ?
    \frac{1}{x^{1/4}+x}=\frac{1}{x^{1/4} \left(1+x^{3/4}\right)}

    Now, substitue y=x^{3/4}

    dy=\frac 34 \cdot x^{-1/4} ~dx \implies dx=\frac 43 \cdot \frac{dy}{x^{-1/4}}

    Going back to the integral :

    2 \int \frac{1}{x^{1/4} \left(1+x^{3/4}\right)}=2 \int \frac{1}{{\color{red}x^{1/4}} \left(1+y\right)} \cdot \frac 43 \cdot \frac{1}{\color{red}x^{-1/4}} ~dy


    Do you see any simplification with the red parts ??



    Edit : you could have substituted y=1+x^{3/4}, but it doesn't matter, you got it
    Last edited by Moo; September 23rd 2008 at 01:25 PM.
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  8. #8
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    yep! it cancels to just x
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  9. #9
    Moo
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    Quote Originally Posted by plooms View Post
    yep! it cancels to just x
    No, it cancels to 1

    x^a \cdot x^{-a}=x^{a-a}=x^0=1

    Can you solve the integral now ?

    2 \times \frac 43 \int  \frac{1}{1+y} ~dy
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  10. #10
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    ohhh durr!

    1/y would integrate to log(y)

    (8/3)[log(x^(3/4))+1]+c ?
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  11. #11
    Moo
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    Quote Originally Posted by plooms View Post
    ohhh durr!

    1/y would integrate to log(y)

    (8/3)[log(x^(3/4)+1)]+c ?
    The 1 is in the logarithm

    \frac{1}{1+y} integrates into \ln(1+y)

    See ? Not that bad !
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  12. #12
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    thanks a lot for the detailed responses!
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  13. #13
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    Quote Originally Posted by plooms View Post

    Integrate 2/[x^(1/4)+x]

    How would i go about finding a substitution for that?
    Put x=z^4 and the integral becomes 8\int{\frac{z^{3}}{z+z^{4}}\,dz}=8\int{\frac{z^{2}  }{1+z^{3}}\,dz}=\frac{8}{3}\ln \left| 1+z^{3} \right|+k.

    Back substitute and we're done.
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