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Math Help - Integrals

  1. #1
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    Integrals


    and

    Are the only ones left in my assignment, please help me if you can, i haven't got much time. Thank you
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  2. #2
    Behold, the power of SARDINES!
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    I love this proceedure

    We will use integration by parts twice and then solve for the integral

    use u=e^{3x} \mbox{ and }dv=\sin(x)

    \int e^3x \sin(x)dx=-e^{3x} \cos(x)+3 \int e^{3x} \cos(x)dx

    Now we will integrate by parts again with

    u=e^{3x} \mbox{ and } dv=\cos(x)

    \int e^3x \sin(x)dx=-e^{3x} \cos(x)+3[e^{3x} \sin(x) -3 \int e^{3x}\sin(x)dx]

    distributing and simplifying we get

    \int e^3x \sin(x)dx=-e^{3x} \cos(x)+3e^{3x} \sin(x) -9 \int e^{3x}\sin(x)dx

    now we adding 9 \int e^{3x}\sin(x)dx to BOTH sides of the equation.

    10\int e^3x \sin(x)dx=-e^{3x} \cos(x)+3e^{3x} \sin(x)

    Solving for the integral we get...

    \int e^{3x}\sin(x)dx=\frac{e^{3x}}{10}\left( 3\sin(x)-\cos(x)\right)
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  3. #3
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    Thank you so much, and btw, the second task, which i posted, is not the one i need to solve, i solved that myself. This is the one:
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  4. #4
    Behold, the power of SARDINES!
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    \int \frac{\cos(3x)}{\sqrt{8+\cos^2(3x)}}dx

    Note that the denominator can be rewritten as follows

    8+\cos^2(3x)=8 + (1-1)+\cos^2(3x)=9+\cos^2(3x)-1=9-\sin^2(3x)

    so now we have the integral

    \int \frac{\cos(3x)}{\sqrt{9-\sin^2(3x)}}dx

    now using u=\frac{1}{9}\sin(3x) \mbox{ and } du=\frac{1}{3}\cos(3x)dx

    \int \frac{3du}{\sqrt{9-9u^2}}=\int \frac{du}{\sqrt{1-u^2}}

    You should be able to finish from here.

    Good luck
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  5. #5
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    u=2+\sin^3(x) \mbox{ and } du=3\sin^2(x)\cos(x)dx

    \int \frac{\cos(x) \sin^2(x)}{\sqrt[3]{2+sin^3(x)}}dx=\frac{1}{3}\int u^{-1/3}du = \frac{1}{2}u^{2/3}

    =\frac{1}{2} \sqrt[3]{(2+sin^3(x))^2}

    P.s if you have new questions post them in a new thread.

    Last edited by TheEmptySet; March 27th 2008 at 01:12 AM. Reason: lost a 3 :(
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