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Math Help - need some calrification on integrating trig functions

  1. #1
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    need some calrification on integrating trig functions

    hi peeps..
    there is something bothering me about a very simple integral where 2 easily justified methods just barely dont reach the same result.

    suppose there is given an integral of (2cos(x)sin(x))... one method involving using the trig identities (Double Angle Formula) would be to first manipulate "2cos(x)sin(x)" to sin(2x) then easily solve the integral of that..

    by doing that we see that integral[2cos(x)sin(x)dx] = integral[sin(2x)dx] =
    -(1/2)cos(2x)...

    OK, now here comes the method that i had chosen to use priorly and all i did was take the '2' outside of the integral and substitute u=sin(x) in order to eliminate cos(x) as shown below..

    Preliminary steps {u=sin(x), then du=cos(x)dx, then dx=du/cos(x)}

    integral[2cos(x)sin(x)dx] = 2integral[cos(x)sin(x)dx] = 2integral[cos(x)u du/cos(x)] = 2integral[u du] = 2[(1/2)u^2] = sin(x)^2...

    sin(x)^2 is 1/2(1-cos(2x)) and that does not match -(1/2)cos(2x) as highlight above. and its just a difference of just a 1/2 factor!

    did i do anything wrong? i was just assuming that both ways should reach to the same result..
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  2. #2
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    In both cases you left out

    + a constant.

    Should I say more?


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  3. #3
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    hey

    ohhh... so that was the misconception...
    so that extra half couldve been regarded as part of a constant?

    so you're telling me that
    1/2 - 1/2cos(2x) + C
    and
    -1/2cos(2x) + C
    are the same things??

    plzzz reply
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  4. #4
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    Quote Originally Posted by ramzouzy View Post
    ohhh... so that was the misconception...
    so that extra half couldve been regarded as part of a constant?

    so you're telling me that
    1/2 - 1/2cos(2x) + C
    and
    -1/2cos(2x) + C
    are the same things??

    plzzz reply
    Yes you've got it.
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