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Thread: Intergration

  1. #1
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    Red face Intergration

    This problem I cannot seem to solve,I have the answer but i want to know how to get to it, i do not know if i used the table or do i integrate it by parts? Thank You!!
    The problem:
    Integrate[Sin[3 x] Cos[5 x], x]

    and this is the answer:

    8 (Cos[x]) ^(2)- Cos[8 x]
    -------------------------
    16
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  2. #2
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    pickslides's Avatar
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    Make

    $\displaystyle \int \sin(3x)\cos(5x)~dx$

    into

    $\displaystyle \int \sin(4x-x)\cos(4x+x)~dx$

    and using Sum-Difference formulas from here

    Table of Trigonometric Identities

    $\displaystyle \int \sin(4x-x)\cos(4x+x)~dx= $ $\displaystyle \int (\sin(4x)\sin(x)-\cos(4x)\cos(x))(\cos(4x)\cos(x)-\sin(4x)\sin(x))~dx$

    now simplify and integrate.
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  3. #3
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    Quote Originally Posted by pickslides View Post
    Make

    $\displaystyle \int \sin(3x)\cos(5x)~dx$

    into

    $\displaystyle \int \sin(4x-x)\cos(4x+x)~dx$

    and using Sum-Difference formulas from here

    Table of Trigonometric Identities

    $\displaystyle \int \sin(4x-x)\cos(4x+x)~dx= $ $\displaystyle \int (\sin(4x)\sin(x)-\cos(4x)\cos(x))(\cos(4x)\cos(x)-\sin(4x)\sin(x))~dx$

    now simplify and integrate.
    at the end when you plug back in the trig identities, the sum-difference,from sin(4x-x) is not right
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  4. #4
    Super Member redsoxfan325's Avatar
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    $\displaystyle \sin(4x-x)=\sin(4x)\cos(x)-\cos(4x)\sin(x)$
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  5. #5
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by flutterby View Post
    This problem I cannot seem to solve,I have the answer but i want to know how to get to it, i do not know if i used the table or do i integrate it by parts? Thank You!!
    The problem:
    Integrate[Sin[3 x] Cos[5 x], x]

    and this is the answer:

    8 (Cos[x]) ^(2)- Cos[8 x]
    -------------------------
    16
    You can also do this using the "cyclic method". (That's not it's official name; that's just what I call it.)

    Let $\displaystyle I=\int\sin3x\cos5x\,dx$

    $\displaystyle u=\sin3x$ and $\displaystyle dv=\cos5x\,dx$
    $\displaystyle du=3\cos3x\,dx$ and $\displaystyle v=\frac{1}{5}\sin5x$

    The integration by parts formula gives:

    $\displaystyle I=\frac{1}{5}\sin3x\sin5x-\frac{3}{5}\int\cos3x\sin5x\,dx$

    $\displaystyle u=\cos3x$ and $\displaystyle dv=\sin5x\,dx$
    $\displaystyle du=-3\sin3x\,dx$ and $\displaystyle v=-\frac{1}{5}\cos5x$

    Now we have:

    $\displaystyle I=\frac{1}{5}\sin3x\sin5x-\frac{3}{5}\int\cos3x\sin5x\,dx$ $\displaystyle =\frac{1}{5}\sin3x\sin5x-\frac{3}{5}\left[-\frac{1}{5}\cos3x\cos5x-\frac{3}{5}\int\sin3x\cos5x\,dx\right]$

    Note that that last integral is just $\displaystyle I$, so we have:

    $\displaystyle I=\frac{1}{5}\sin3x\sin5x-\frac{3}{5}\left[-\frac{1}{5}\cos3x\cos5x-\frac{3}{5}I\right]$

    Solve for $\displaystyle I$. (You should also check my math; make sure I didn't make any sign errors and whatnot.)

    Remark: To get the answer into the format given to you in the answer key, you may need to use some trig identities.
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