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Thread: Some Antideriv-integral problems

  1. #1
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    Some Antideriv-integral problems

    Find the antiderivatives - I tired working on these but I can't seem to get the correct answer for it
    1. $\displaystyle
    \int e^-^1 \cos (3t) dt
    $
    2. $\displaystyle
    \int \sin^3x \sqrt {cos (x)} dx
    $ ----- Nevermind, I got this one
    3. $\displaystyle
    \int^{4}_{1} \sqrt {t} \ln(t) dt
    $
    4. $\displaystyle
    \int^{\pi/4}_{0} \sin (3x) \cos (x) dx
    $
    Last edited by nirva; May 20th 2006 at 04:25 PM.
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  2. #2
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    Quote Originally Posted by nirva
    3. $\displaystyle
    \int^{4}_{1} \sqrt {t} \ln(t) dt
    $
    Using integration by parts by allowing,
    $\displaystyle u=\ln t$
    $\displaystyle v'=\sqrt{t}$
    Thus,
    $\displaystyle u'=1/t=t^{-1}$
    $\displaystyle v=2/3t^{3/2}$
    Thus, we have,
    $\displaystyle \frac{2}{3} \ln t \cdot t^{3/2}-\frac{2}{3}\int t^{3/2}t^{-1}dt$
    Thus, the integrand is,
    $\displaystyle \int t^{1/2}dt=\frac{2}{3}t^{3/2}$
    Thus,
    $\displaystyle \left \frac{2}{3} \ln t \cdot t^{3/2}-\frac{4}{9}t^{3/2} \right|^{4}_1 \approx 4.28$
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  3. #3
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    Quote Originally Posted by nirva
    4. $\displaystyle
    \int^{\pi/4}_{0} \sin (3x) \cos (x) dx
    $
    Use integration by parts with,
    $\displaystyle u=\sin 3x$
    $\displaystyle v'=\cos x$
    Thus,
    $\displaystyle u'=3\cos 3x$
    $\displaystyle v=\sin x$
    The integrand becomes,
    $\displaystyle \sin 3x\sin x-3\int \cos 3x\sin xdx$
    Use integration by parts again with,
    $\displaystyle u=\cos 3x$
    $\displaystyle v'=\sin x$
    Thus,
    $\displaystyle u'=-3\sin 3x$
    $\displaystyle v=-\cos x$
    Thus, the integrand becomes,
    $\displaystyle \sin 3x\sin x-3 \left( -\cos 3x \cos x-3\int \sin 3x\cos x dx \right)$
    Thus,
    $\displaystyle \int \sin (3x) \cos (x) dx$=$\displaystyle \sin 3x\sin x+3\cos 3x\cos x+9\int \sin 3x \cos xdx$
    Thus, (solve for this integral as if a variable)
    $\displaystyle -\frac{1}{8}\left( \sin 3x\sin x+3\cos 3x\cos x \right)$
    Since you know the anti-derivative you can find the definite integral.
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  4. #4
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    Quote Originally Posted by nirva
    Find the antiderivatives - I tired working on these but I can't seem to get the correct answer for it
    1. $\displaystyle
    \int e^-^1 \cos (3t) dt
    $
    I presume you mean,
    $\displaystyle \int e^{-t} \cos 3t dt$
    Use integration by parts with,
    $\displaystyle u'=e^{-t}$
    $\displaystyle v=\cos 3t$
    Thus,
    $\displaystyle u=-e^{-t}$
    $\displaystyle v'=-3\sin 3t$
    By parts,
    $\displaystyle -e^{-t}\cos 3t-3\int e^{-t}\sin 3t dt$
    Use integration by parts again,
    $\displaystyle u'=e^{-t}$
    $\displaystyle v=\sin 3t$
    Thus,
    $\displaystyle u=-e^{-t}$
    $\displaystyle v'=3\cos 3t$
    By parts,
    $\displaystyle -e^{-t}\cos 3t-3\left(-e^{-t}\sin 3t+3\int e^{-t}\cos 3t dt \right)$
    Thus,
    $\displaystyle \int e^{-t} \cos 3t dt$=$\displaystyle -e^{-t}\cos 3t+3e^{-t}\sin 3t-9\int e^{-t}\cos 3tdt$
    Solve for integrand as if to say an equation,
    $\displaystyle \int e^{-t} \cos 3t dt=\frac{1}{10}e^{-t}(3\sin 3t-\cos t)+C$
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  5. #5
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    Quote Originally Posted by ThePerfectHacker
    I presume you mean,
    $\displaystyle \int e^{-t} \cos 3t dt$
    Use integration by parts with,
    $\displaystyle u'=e^{-t}$
    $\displaystyle v=\cos 3t$
    Thus,
    $\displaystyle u=-e^{-t}$
    $\displaystyle v'=-3\sin 3t$
    Yea it is $\displaystyle \int e^{-t} \cos 3t dt$ The t was too small, sorry
    Is the lable for the integration by parts wrong? Like the u being the u'
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  6. #6
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    Quote Originally Posted by nirva
    Yea it is $\displaystyle \int e^{-t} \cos 3t dt$ The t was too small, sorry
    Is the lable for the integration by parts wrong? Like the u being the u'
    Let me give you a tip for LaTeX,
    If you want to write,
    $\displaystyle e^{-t}$
    I saw you did it as,
    Code:
    e^-^t
    The more advaned way is to write,
    Code:
    e^{-t}
    Note the { } play a role of parantheses in LaTeX.
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