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Thread: Integration Question help please

  1. #1
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    Integration Question help please

    ----------------------------------------------------------
    Integrate x^(1/2) / (1 + x) dx
    -----------------------------------
    I substituted x^(1/2) = sin y

    so it becomes Integral of 2*tany*siny dy

    then integration by parts its

    2*(tany*cosy - Integral of (-secy) dy)

    = 2ln| sec y + tan y | - 2 siny

    then substituting x back in

    = 2ln| (1 + x^(1/2)) / (1 - x)^(1/2) | - 2x^(1/2)

    but the answer in the back of the book is

    2x^(1/2) - 2tan^-1( x^(1/2) )

    If anyone can tell me what I'm doing wrong it will be much appreciated
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  2. #2
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    Divide top and bottom (of the integrand) by root x, then re-write the top (now x) as 1 + x -1. Will add a pic...

    Edit:

    Just in case a picture helps (same as redsox below, I guess)...



    ... where



    ... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (which is the inner function of the composite, subject to the chain rule).


    __________________________________________

    Don't integrate - balloontegrate!

    http://www.ballooncalculus.org/forum/top.php

    Draw balloons with LaTeX: http://www.ballooncalculus.org/asy/doc.html
    Last edited by tom@ballooncalculus; Sep 15th 2009 at 08:24 AM. Reason: pic
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  3. #3
    MHF Contributor Calculus26's Avatar
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    you wan to let sqrt(x) = tan(y)

    I think you mistakenly used 1 + sin^2(y) = cos^2(y) in the denominator


    but 1+ sin^2(y) = 2 - cos^2(y)

    See attachment
    Attached Thumbnails Attached Thumbnails Integration Question help please-sqrtingrl.jpg  
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  4. #4
    MHF Contributor Calculus26's Avatar
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    the next to last line in attachment should read

    sqrt(x)dx/(1+x) = 2sec^2(y) -2
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  5. #5
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by kinetix2006 View Post
    ----------------------------------------------------------
    Integrate x^(1/2) / (1 + x) dx
    -----------------------------------
    I substituted x^(1/2) = sin y

    so it becomes Integral of 2*tany*siny dy

    then integration by parts its

    2*(tany*cosy - Integral of (-secy) dy)

    = 2ln| sec y + tan y | - 2 siny

    then substituting x back in

    = 2ln| (1 + x^(1/2)) / (1 - x)^(1/2) | - 2x^(1/2)

    but the answer in the back of the book is

    2x^(1/2) - 2tan^-1( x^(1/2) )

    If anyone can tell me what I'm doing wrong it will be much appreciated
    \int\frac{\sqrt{x}}{1+x}\,dx

    I would do it like this (to avoid using trig terms, which can get messy):

    Let x=u^2. Thus dx=2u\,du.

    Now you have \int\frac{2u^2}{1+u^2}\,du = \int\frac{2u^2+2-2}{1+u^2}\,du =  \int\frac{2u^2+2}{u^2+1}-\frac{2}{1+u^2}\,du = \int 2-2\frac{1}{1+u^2}\,du = 2u-2\arctan(u)

    Now sub back in u=\sqrt{x} and add C and you're done.
    <br />
\boxed{\int\frac{\sqrt{x}}{1+x}\,dx = 2\sqrt{x}-2\arctan(\sqrt{x})+C}
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