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Math Help - 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.  <br />
\int e^-^1 \cos (3t) dt<br />
    2. <br />
\int \sin^3x \sqrt {cos (x)} dx<br />
----- Nevermind, I got this one
    3. <br />
\int^{4}_{1} \sqrt {t} \ln(t) dt<br />
    4. <br />
\int^{\pi/4}_{0} \sin (3x) \cos (x) dx<br />
    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. <br />
\int^{4}_{1} \sqrt {t} \ln(t) dt<br />
    Using integration by parts by allowing,
    u=\ln t
    v'=\sqrt{t}
    Thus,
    u'=1/t=t^{-1}
    v=2/3t^{3/2}
    Thus, we have,
    \frac{2}{3} \ln t \cdot t^{3/2}-\frac{2}{3}\int t^{3/2}t^{-1}dt
    Thus, the integrand is,
    \int t^{1/2}dt=\frac{2}{3}t^{3/2}
    Thus,
     \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. <br />
\int^{\pi/4}_{0} \sin (3x) \cos (x) dx<br />
    Use integration by parts with,
    u=\sin 3x
    v'=\cos x
    Thus,
    u'=3\cos 3x
    v=\sin x
    The integrand becomes,
    \sin 3x\sin x-3\int \cos 3x\sin xdx
    Use integration by parts again with,
    u=\cos 3x
    v'=\sin x
    Thus,
    u'=-3\sin 3x
    v=-\cos x
    Thus, the integrand becomes,
    \sin 3x\sin x-3 \left( -\cos 3x \cos x-3\int \sin 3x\cos x dx \right)
    Thus,
    \int \sin (3x) \cos (x) dx= \sin 3x\sin x+3\cos 3x\cos x+9\int \sin 3x \cos xdx
    Thus, (solve for this integral as if a variable)
    -\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.  <br />
\int e^-^1 \cos (3t) dt<br />
    I presume you mean,
    \int e^{-t}  \cos 3t dt
    Use integration by parts with,
    u'=e^{-t}
    v=\cos 3t
    Thus,
    u=-e^{-t}
    v'=-3\sin 3t
    By parts,
    -e^{-t}\cos 3t-3\int e^{-t}\sin 3t dt
    Use integration by parts again,
    u'=e^{-t}
    v=\sin 3t
    Thus,
    u=-e^{-t}
    v'=3\cos 3t
    By parts,
    -e^{-t}\cos 3t-3\left(-e^{-t}\sin 3t+3\int e^{-t}\cos 3t dt \right)
    Thus,
    \int e^{-t} \cos 3t dt= -e^{-t}\cos 3t+3e^{-t}\sin 3t-9\int e^{-t}\cos 3tdt
    Solve for integrand as if to say an equation,
    \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,
    \int e^{-t}  \cos 3t dt
    Use integration by parts with,
    u'=e^{-t}
    v=\cos 3t
    Thus,
    u=-e^{-t}
    v'=-3\sin 3t
    Yea it is \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 \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,
    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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