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Math Help - intergration

  1. #1
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    intergration

    antidifferentiation of

    (x^3)/squareroot((3x^4)+5) dx

    i know the formula 1/square root(1 - x^2) but that doesnt help me.


    can some one please check this one for me

    antidifferentiation

    sin^2xcos^2x dx

    i got 1/3sinx^3 - 1/2sinx^4 + 1/5sin^5 + C

    not very confident with this answer

    also could some one check the antidifferentiation of

    sin3xcos4x dx

    i got 1/4sinx - 1/14sinx + C

    thanks in advance
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  2. #2
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    Quote Originally Posted by helmszee View Post
    antidifferentiation of

    (x^3)/squareroot((3x^4)+5) dx
    \int\frac{x^3}{\sqrt{3x^4+5}}\,dx

    All you need to do is set u=\sqrt{3x^4+5}

    Quote Originally Posted by helmszee View Post
    antidifferentiation

    sin^2xcos^2x dx
    \int\sin^2x\cos^2x\,dx=\frac14\int\sin^22x

    Now define u=2x, for the remaining integral use the formula \cos2v=\cos^2v-\sin^2v.

    Quote Originally Posted by helmszee View Post
    also could some one check the antidifferentiation of

    sin3xcos4x dx
    From

    \sin (\alpha + \beta ) = \sin \alpha \cos \beta + \cos \alpha \sin \beta

    and \sin (\alpha - \beta ) = \sin \alpha \cos \beta - \cos \alpha \sin \beta

    Derive a formula for \sin \alpha \cos \beta
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  3. #3
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    Quote Originally Posted by helmszee View Post
    antidifferentiation of

    (x^3)/squareroot((3x^4)+5) dx

    i know the formula 1/square root(1 - x^2) but that doesnt help me.
    Observe \frac{d}{dx}~\sqrt{3x^4+5}=\frac{(1/2)(12x^3)}{\sqrt{3x^4+5}}.

    Therefore:

    <br />
\int \frac{x^3}{\sqrt{3 x^4+5}} ~dx = \frac{1}{6}~\sqrt{3x^4+5}+C<br />

    RonL
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  4. #4
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    Quote Originally Posted by helmszee View Post
    can some one please check this one for me

    antidifferentiation

    sin^2xcos^2x dx

    i got 1/3sinx^3 - 1/2sinx^4 + 1/5sin^5 + C

    not very confident with this answer
    Have you tried differentiating it to see if you get what you started with?

    RonL
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  5. #5
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    antidifferentiation of sin3xcos4x dx

    using sinaxcosbx = 1/2sin(a-b)x + 1/2sin(a+b)x
    =1/2sin(3-4)x + 1/2sin(3+4)x

    -1/2sinx + 1/2sin7x dx

    1/2cosx - 1/14cos7x + C

    Which can't be simplified... i dont think..

    as for antifdifferentiation of sin^2xcos^2x dx

    let u = sinx
    so du = cosx dx

    antidif sin^2xcos^2x dx = sin^2xcos^2xcosx dx
    = sin^2x(1-sinx)^2cosx dx
    = u^2(1-u)2 du
    = u^2(1-2u +u^2)
    = (u^2 - 2u^3 + u^4) du
    = 1/3u^3 - 1/2u^4 + 1/5u^5 + C
    = 1/3sinx^3 - 1/2sinx^4 + 1/5sin^5 + C

    hope that makes sence wen ever i used the math wrap text thing it always stuffs it up
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by helmszee View Post
    antidifferentiation of sin3xcos4x dx

    using sinaxcosbx = 1/2sin(a-b)x + 1/2sin(a+b)x
    =1/2sin(3-4)x + 1/2sin(3+4)x

    -1/2sinx + 1/2sin7x dx

    1/2cosx - 1/14cos7x + C

    Which can't be simplified... i dont think..
    that is correct


    as for antifdifferentiation of sin^2xcos^2x dx

    let u = sinx
    so du = cosx dx

    antidif sin^2xcos^2x dx = sin^2xcos^2xcosx dx
    ...
    here is your mistake, the two sides are not equal. you have cos^2(x) on the left but cos^3(x) on the right. follow the suggestion in post #2
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