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Math Help - Integrate this equation

  1. #1
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    Integrate this equation

    I am unable to initiate the first few steps.....
    solve

    \int sin4x - cos6x
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  2. #2
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    Quote Originally Posted by zorro View Post
    I am unable to initiate the first few steps.....
    solve

    \int sin4x - cos6x
    You're expected to know the following standard forms:

    \int \sin (kx) \, dx = - \frac{1}{k} \cos (kx) + C

    \int \cos (kx) \, dx = \frac{1}{k} \sin (kx) + C
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    You're expected to know the following standard forms:

    \int \sin (kx) \, dx = - \frac{1}{k} \cos (kx) + C

    \int \cos (kx) \, dx = \frac{1}{k} \sin (kx) + C
    I am using a different technique please let me know if it is right or no...
    I = I_1 + I_2
    I_1 = \int sin(4x)dx
    4x=u_1
    4=\frac{du_1}{dx}
    Therefore \ I_1=\frac{1}{4} \int 4 sin u_1 du_1
    I_1= \frac{1}{4} cos u_1 +c_1


    I_2 = \int cos(6x)dx
    \frac{1}{6} \int 6 cos(6x)dx
    6x=u_2
    6=\frac{du_2}{dx}
    I_2=\frac{1}{6} \int cosu_2 du_2
    I_2=\frac{-sin6x}{6}+c_2

    Therefore
    <br />
I=I_1+I_2 <br />
=\frac{cos4x}{4} - \frac{sin6x}{6} +c<br />
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  4. #4
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    Quote Originally Posted by zorro View Post
    I am using a different technique please let me know if it is right or no...
    I = I_1 + I_2
    I_1 = \int sin(4x)dx
    4x=u_1
    4=\frac{du_1}{dx}
    Therefore \ I_1=\frac{1}{4} \int 4 sin u_1 du_1
    I_1= {\color{red} - } \frac{1}{4} cos u_1 +c_1


    I_2 = \int cos(6x)dx
    \frac{1}{6} \int 6 cos(6x)dx
    6x=u_2
    6=\frac{du_2}{dx}
    I_2=\frac{1}{6} \int cosu_2 du_2
    I_2=\frac{-sin6x}{6}+c_2

    Therefore
    <br />
I=I_1+I_2 <br />
= {\color{red} - } \frac{cos4x}{4} - \frac{sin6x}{6} +c<br />
    Note the corrections in red.
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