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Math Help - integration using trigonmetric functions

  1. #1
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    integration using trigonmetric functions

    integrate question:

    <br /> <br />
= \int (3sin(4x+5) - 6cos(7-8x)) dx<br /> <br />

    any help, im thinking it has to do with identities?
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  2. #2
    Newbie DJMinkus's Avatar
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    It looks kind of evil, but it's actually not that difficult. You can split it up into two different integrals. 3\int (sin(4x+5))dx - 6\int (cos(7-8x))dx

    Now use u-subs, taking the 4x+5 and the 7-8x to be the u's for the two different integrals. I think it yields:
    (-3/4)cos(4x+5) +(3/4)sin(7-8x) + C
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  3. #3
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    Quote Originally Posted by DJMinkus View Post
    It looks kind of evil, but it's actually not that difficult. You can split it up into two different integrals. 3\int (sin(4x+5))dx - 6\int (cos(7-8x))dx

    Now use u-subs, taking the 4x+5 and the 7-8x to be the u's for the two different integrals. I think it yields:
    (-3/4)cos(4x+5) +(3/4)sin(7-8x) + C
    shouldnt it be (-3/4)cos(4x+5)-(3/4)sin(7-8x) + C ? if theres a minus there already and u have a plus from the sin coming in for the cos?
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  4. #4
    Newbie DJMinkus's Avatar
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    But don't forget, the coefficient for x in the sin function is -8, making du = -8 and thus causing the two minuses (from the original equation and then from the integration) to cancel out, yielding a positive.
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