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Math Help - need to check if this is right

  1. #1
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    need to check if this is right

    have to integrate this
    (3cos4x-2sec^2(3x)+tan x) dx

    now i get
    3sin4x-2tan3x-3x+ln secx
    is this right or do i have to treat each part as like a function of a function
    and say like for the first bit
    3=U and cos4x=V
    and then use du and dv and work it all out then put it back togther
    ty
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  2. #2
    MHF Contributor Amer's Avatar
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    Jordan
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    Quote Originally Posted by alpharebel10 View Post
    have to integrate this
    (3cos4x-2sec^2(3x)+tan x) dx

    now i get
    3sin4x-2tan3x-3x+ln secx
    is this right or do i have to treat each part as like a function of a function
    and say like for the first bit
    3=U and cos4x=V
    and then use du and dv and work it all out then put it back togther
    ty
    in first one and the second see this you will find your mistake

    \int sin(6x)dx=\frac{-cos(6x)}{6}+c

    the final one is
    \int \frac{f'(x)}{f(x)} = ln(f(x)) +c

    do you see your mistakes
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  3. #3
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    Jan 2009
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    thanks for the help

    can u just check if this is right i hope it is ty
    int 3cos4x-2sec^2(3x)+tanx

    frac{3sin4x}{4}-frac{2tan3x}{3}+-lncosx
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  4. #4
    MHF Contributor Amer's Avatar
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    Jordan
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    Quote Originally Posted by alpharebel10 View Post
    can u just check if this is right i hope it is ty
    int 3cos4x-2sec^2(3x)+tanx

    \frac{3sin4x}{4}-\frac{2tan3x}{3}+-lncosx

    [tex]\frac{3sin4x}{4}-\frac{2tan3x}{3}+-lncosx[ /math]

    [tex]\int 3cos4x-2sec^2(3x)+tanx[ /math]

    you solution is correct but you forget to put \ before the frac to present your solution in latex
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