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Math Help - 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

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

    into

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

    and using Sum-Difference formulas from here

    Table of Trigonometric Identities

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

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

    into

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

    and using Sum-Difference formulas from here

    Table of Trigonometric Identities

     \int \sin(4x-x)\cos(4x+x)~dx= \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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    \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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    Swampscott, MA
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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 I=\int\sin3x\cos5x\,dx

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

    The integration by parts formula gives:

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

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

    Now we have:

    I=\frac{1}{5}\sin3x\sin5x-\frac{3}{5}\int\cos3x\sin5x\,dx =\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 I, so we have:

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

    Solve for 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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