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Math Help - integration problems

  1. #1
    Junior Member scottie.mcdonald's Avatar
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    integration problems

    i'm having trouble (as always), starting these problems. hints on how to start would be nice.

    evaluate the integrals

    a). \int sin\sqrt(at)dt

    i think i have to do this by integration by parts, so i have:

     f(x) = sin \sqrt(at), f\prime(x) = -cos \sqrt(at), g(x)=t, g\prime(x)=dt .

    though i think this is wrong.

    b).  \int sin4xcos3xdx .

    again, i think it's integration by parts, but the 4 and 3 are messing me up

    c).  \int \sqrt(1+e^x)dx

    i have  f(x)=1+e^x, f\prime(x)=xe^x, g(x)=x, g\prime(x)=dx

    the square root is messing me up, so again, i don't think i selected the parts correctly.

    any help on how to get started would be nice.

    thank you,

    Scott
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  2. #2
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    Hello, scottie.mcdonald!

    The third one requires a "simple" substitution . . .


    c)\;\;\int \sqrt{1+e^x}\,dx

    Let: . u \:=\:\sqrt{1+e^x} \quad\Rightarrow\quad u^2 \:=\:1 + e^x \quad\Rightarrow\quad e^x \:=\:u^2-1 \quad\Rightarrow\quad x \:=\:\ln(u^2-1)

    . . Then: . dx \:=\:\frac{2u}{u^2-1}\,du


    Substitute: . \int u \left(\frac{2u}{u^2-1}\,du\right) \;=\;2\int\frac{u^2}{u^2-1}\,du

    . . =\;2\int\left(1 + \frac{1}{u^2-1}\right)\,du  \;=\;2\left[u + \frac{1}{2}\ln\left(\frac{u-1}{u+1}\right)\right] \;+\; C \;=\;2u \;+ \;\ln\left(\frac{u-1}{u+1}\right) \;+\; C


    Back-substitute: . 2\sqrt{1+e^x} + \ln\left(\frac{\sqrt{1+e^x} - 1}{\sqrt{1+e^x} + 1}\right) + C .[1]


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    We could stop here, but we can simplify the answer further . . .

    Multiply the fracton by \frac{\sqrt{1+e^x} -1}{\sqrt{1+e^x}-1}

    . . \frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x} + 1} \cdot \frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x} - 1} \;=\;\frac{(\sqrt{1+e^x}-1)^2}{(1+e^x) - 1} \;=\;\frac{(\sqrt{1+e^x} -1)^2}{e^x}<br />


    Then [1] becomes: . 2\sqrt{1+e^x} + \ln\left[\frac{(\sqrt{1+e^x} - 1)^2}{e^x}\right] + C

    . . . . . . . . . . . . = \;2\sqrt{1+e^x} + \ln(\sqrt{1+e^x} - 1)^2 - \ln(e^x) + C


    . . . . . . . . . . . . = \;2\sqrt{1+e^x} + 2\ln(\sqrt{1+e^x}-1) - x + C


    Pull something like this on any of your exams,
    . . and I'll guarentee that you'll surprise/impress/terrify your teacher.

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  3. #3
    Junior Member scottie.mcdonald's Avatar
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    ha ha yeah, my teacher would look at me with wonder if i thought about rationalizing at the end (though now i'm always going to look for that).

    Thanks for the help =)
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