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Math Help - Integration Question

  1. #1
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    Integration Question

    Hi

    I need help on the following questions:

    1)  \int 5cos(2x) - 10xsin(2x) dx = k(xcos(2x))+C
    This is what i have done.

    5 \int cos(2x) - 2xsin(2x)

    5 \int \frac{sin(2x)}{2} - 2 \int xsin(2x)

    u = x du = 1 dx

    dv = sin(2x) dx v= \frac{-cos(2x)}{2}

    \frac{5}{2}sin(2x) - 2[ \frac{-xcos(2x)}{2} - \int \frac{-cos(2x)}{2}]

    final answer.
    \frac{5}{2}sin(2x) + xcos(2x) - \frac{sin(2x)}{2}

    2) \int sin(2x)sin(5x) dx
    This is what i have done, not sure if it is correct.
    \int \frac{1}{2}cos(2x-5x)-cos(2x+5x)

    \frac{1}{2} \int cos(-3x) - cos(7x)

    \frac{1}{2}[ \frac{-sin(-3x)}{3} - \frac{sin(7x)}{7}]

    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    Hi

    I need help on the following questions:

    1)  \int 5cos(2x) - 10xsin(2x) dx = k(xcos(2x))+C
    This is what i have done.

    5 \int cos(2x) - 2xsin(2x)

    5 \int \frac{sin(2x)}{2} - 2 \int xsin(2x)

    u = x du = 1 dx

    dv = sin(2x) dx v= \frac{-cos(2x)}{2}

    \frac{5}{2}sin(2x) - 2[ \frac{-xcos(2x)}{2} - \int \frac{-cos(2x)}{2}]

    final answer.
    \frac{5}{2}sin(2x) + xcos(2x) - \frac{sin(2x)}{2}

    2) \int sin(2x)sin(5x) dx
    This is what i have done, not sure if it is correct.
    \int \frac{1}{2}cos(2x-5x)-cos(2x+5x)

    \frac{1}{2} \int cos(-3x) - cos(7x)

    \frac{1}{2}[ \frac{-sin(-3x)}{3} - \frac{sin(7x)}{7}]

    P.S
    Question 1 is correct.


    2. \int{\sin{(2x)}\sin{(5x)}\,dx}.

    Using integraton by parts with

    u = \sin{(2x)} so du = 2\cos{(2x)}

    dv = \sin{(5x)} so v = -\frac{1}{5}\cos{(5x)}

    we find

    \int{\sin{(2x)}\sin{(5x)}\,dx} = -\frac{1}{5}\sin{(2x)}\cos{(5x)} - \int{-\frac{2}{5}\cos{(2x)}\cos{(5x)}\,dx}

     = -\frac{1}{5}\sin{(2x)}\cos{(5x)} + \frac{2}{5}\int{\cos{(2x)}\cos{(5x)}\,dx}.


    Now using integration by parts again with

    u = \cos{(2x)} so du = -2\sin{(2x)}

    dv = \cos{(5x)} so v = \frac{1}{5}\sin{(5x)}

    we find

    -\frac{1}{5}\sin{(2x)}\cos{(5x)} +  \frac{2}{5}\int{\cos{(2x)}\cos{(5x)}\,dx}
     = -\frac{1}{5}\sin{(2x)}\cos{(5x)} + \frac{2}{5}\left[\frac{1}{5}\cos{(2x)}\sin{(5x)} - \int{-\frac{2}{5}\sin{(2x)}\sin{(5x)}\,dx}\right]

     = -\frac{1}{5}\sin{(2x)}\cos{(5x)} + \frac{2}{25}\cos{(2x)}\sin{(5x)} + \frac{4}{25}\int{\sin{(2x)}\sin{(5x)}\,dx}



    So we have:

    \int{\sin{(2x)}\sin{(5x)}\,dx}
     = -\frac{1}{5}\sin{(2x)}\cos{(5x)} + \frac{2}{25}\cos{(2x)}\sin{(5x)} +  \frac{4}{25}\int{\sin{(2x)}\sin{(5x)}\,dx}

    \frac{21}{25}\int{\sin{(2x)}\sin{(5x)}\,dx} =  -\frac{1}{5}\sin{(2x)}\cos{(5x)} + \frac{2}{25}\cos{(2x)}\sin{(5x)}

    \int{\sin{(2x)}\sin{(5x)}\,dx} = \frac{25}{21}\left[-\frac{1}{5}\sin{(2x)}\cos{(5x)} + \frac{2}{25}\cos{(2x)}\sin{(5x)}\right]

    \int{\sin{(2x)}\sin{(5x)}\,dx} = -\frac{5}{21}\sin{(2x)}\cos{(5x)} + \frac{2}{21}\cos{(2x)}\sin{(5x)} + C.
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  3. #3
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    Quote Originally Posted by Paymemoney View Post
    Hi

    I need help on the following questions:

    1)  \int 5cos(2x) - 10xsin(2x) dx = k(xcos(2x))+C
    This is what i have done.

    5 \int cos(2x) - 2xsin(2x)

    5 \int \frac{sin(2x)}{2} - 2 \int xsin(2x)
    There should not be an integral sign between "5" and "sin(2x)/2".

    u = x du = 1 dx

    dv = sin(2x) dx v= \frac{-cos(2x)}{2}

    \frac{5}{2}sin(2x) - 2[ \frac{-xcos(2x)}{2} - \int \frac{-cos(2x)}{2}]

    final answer.
    \frac{5}{2}sin(2x) + xcos(2x) - \frac{sin(2x)}{2}

    2) \int sin(2x)sin(5x) dx
    This is what i have done, not sure if it is correct.
    \int \frac{1}{2}cos(2x-5x)-cos(2x+5x)

    \frac{1}{2} \int cos(-3x) - cos(7x)

    \frac{1}{2}[ \frac{-sin(-3x)}{3} - \frac{sin(7x)}{7}]

    P.S
    Yes, this is correct. You can simplify it slightly by writing cos(-3x)= cos(3x) since cosine is an even function. Of course, in your final solution, -sin(-3x)= -(- sin(3x))= sin(3x) since sine is an odd function.

    It would be an interesting exercise in trigonometric identities to show that your solution is the same as Prove It's solution (I did not check his work). Although your solution is considerably easier than his!
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