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Math Help - Help with integral using ln

  1. #1
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    Help with integral using ln

    Well, i was assigned to do this problem but i haven't yet covered the ln formula and my professor still wants me to do it.
    Can someone please show me how to solve this particular integration (See Attachment)?

    Thank you.
    Attached Thumbnails Attached Thumbnails Help with integral using ln-integral-2-.jpg  
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  2. #2
    MHF Contributor Amer's Avatar
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    \int\frac{1}{\sqrt{x+\sqrt{x}}}dx

    let ..x=u^4...dx=4u^3

    \int\frac{1}{\sqrt{(\sqrt{x}\sqrt{x}+\sqrt{x}}}dx

    \int\frac{1}{\sqrt{(\sqrt{x}(\sqrt{x}+1))}}dx

    \int\frac{1}{x^{\frac{1}{4}}\sqrt{(\sqrt{x}+1)}}dx

    now use the substitute

    \int\frac{4u^3}{u\sqrt{(u^2+1)}}du

    \int\frac{4(u^2)}{\sqrt{(u^2+1)}}du

    now..let...tany=u... sec^2y dy=du...secy=\sqrt{1+u^2}

    4u^2=4(u^2+1-1)=4(sec^y-1)=4tan^2u...
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  3. #3
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    Thank you so very much
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  4. #4
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    One last thing.

    So is the final answer (4(tan^2)y)/secy ?
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  5. #5
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Amer View Post
    [tex]

    \int\frac{4u^3}{u\sqrt{(u^2+1)}}du

    \int\frac{4(u^2)}{\sqrt{(u^2+1)}}du

    now..let...tany=u... {\color{red}{sec^2y dy=du}}...secy=\sqrt{1+u^2}

    4u^2=4(u^2+1-1)=4(sec^2y-1)=4tan^2u...
    \int\left(\frac{4tan^2y}{secy}\right)(sec^2y)dy
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  6. #6
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    So after that I solve it like a normal integration or is that how it stays?
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  7. #7
    MHF Contributor Amer's Avatar
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    no you can solve it

    \int\left(\frac{4tan^2y}{secy}\right)(sec^2y)dy

    \int4tan^2y(secy)dy ok

    4\int(sec^2y-1)(secy)dy

    4\int(sec^3y-secy)dy

    4\left(\int sec^3ydy - \int secydy\right)

    do you want me to continue or you can do that...
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  8. #8
    MHF Contributor Amer's Avatar
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    ok

    \int sec^3y dy - \int secy dy

    I want to start with integral on the left
    first multiply the denominator and numerator with sec(y)+tan(y)

    \int secy\left(\frac{secy+tany}{secy+tany}\right) dy

    \int \frac{sec^2y+secy(tany)}{secy+tany} dy

    now you have the numerator is the derive of the denominator so the integral is ln(denominator) since the integrate of f'(x)/f(x) is ln(f(x)

    this is the first
    now

    \int sec^3y dy

    let dv=sec^2y and u=secy
    v=tany.....du=secy tany

    \int sec^3y dy=secy(tany)-\int secy(tan2^y)dy

    \int sec^3y dy=secy(tany)-\int secy(sec^2y-1)dy

    \int sec^3y dy=secy(tany)-\left(\int sec^3y dy-\int secy dy)\right)

    \int sec^3y dy=secy(tany)-\int sec^3y dy+\int secy dy

    2\int sec^3y dy = secy(tany)+\int secy dy

    \int sec^3y dy=\frac{secy(tany)+\int secy dy}{2}

    and the integral in the right we integrate it before
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