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Thread: Integrate this equation

  1. #1
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    Integrate this equation

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

    $\displaystyle \int sin4x - cos6x$
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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

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

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

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

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

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


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

    Therefore
    $\displaystyle
    I=I_1+I_2
    =\frac{cos4x}{4} - \frac{sin6x}{6} +c
    $
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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...
    $\displaystyle I = I_1 + I_2$
    $\displaystyle I_1 = \int sin(4x)dx$
    $\displaystyle 4x=u_1$
    $\displaystyle 4=\frac{du_1}{dx}$
    $\displaystyle Therefore \ I_1=\frac{1}{4} \int 4 sin u_1 du_1$
    $\displaystyle I_1= {\color{red} - } \frac{1}{4} cos u_1 +c_1$


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

    Therefore
    $\displaystyle
    I=I_1+I_2
    = {\color{red} - } \frac{cos4x}{4} - \frac{sin6x}{6} +c
    $
    Note the corrections in red.
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