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Thread: 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

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

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

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

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

    $\displaystyle frac{3sin4x}{4}-frac{2tan3x}{3}+-lncosx$
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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
    $\displaystyle int 3cos4x-2sec^2(3x)+tanx$

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

    [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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