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Math Help - Which two answers are correct?

  1. #1
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    Which two answers are correct?

    Here's the basic question:

    We have an integral, that of which we do not know, where two out of three students got the correct answer. Which two students are right?

    a. sin^2x+C
    b. -cos^2x+C
    c. -sin^2x+C


    My logic is that one of the (-)sin^2x+C is right, meaning the integral might be \int2sin(x)cos(x), but I'm not sure how one could get a -cos^2x+C out of that.

    Any help is appreciated.
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  2. #2
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    Krizalid's Avatar
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    An integral may have many primitives.

    Why don't you differentiate the options to check if they yield the integrand.
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  3. #3
    Senior Member DivideBy0's Avatar
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    Quote Originally Posted by Krizalid View Post
    An integral may have many primitives.

    Why don't you differentiate the options to check if they yield the integrand.
    When you talk about primitives, do you mean a given integral may have two answers which don't actually equal each other but are nonetheless valid evaluations of the integral?
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  4. #4
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    Hello, ebonyscythe!

    This is a classic problem . . .


    Integrate: . 2\int\sin x\cos x\,dx
    There are (at least) three solutions . . .


    (1) Let u = \sin x\quad\Rightarrow\quad du = \cos x\,dx

    Substitute: . 2\int u\,du \;=\;u^2 + C\;=\;\boxed{\sin^2\!x + C}


    (2) Let u = \cos x\quad\Rightarrow\quad du = -\sin x\,dx

    Substitute: . 2\int u(-du) \:=\:-u^2 + C\:=\:\boxed{-\cos^2\!x + C}


    (3) . \sin x\cos x \:=\:\frac{1}{2}(2\sin x\cos x) \;=\;\frac{1}{2}\sin2x

    So we have: . 2\int\frac{1}{2}\sin2x\,dx \;=\;\int\sin2x\,dx \;=\;\boxed{-\frac{1}{2}\cos2x + C}


    . . . . . and these three answers are equivalent.

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  5. #5
    MHF Contributor kalagota's Avatar
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    if the choices were the answers of the 3 students and only two got the correct answer, then student who didn't get it right might had written -\sin^2 x + C, like what kriz wrote, differentiate the three, since we don't really know what the integral was, at least, when the three were differentiated, 2 of them would yield the same derivative..
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  6. #6
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by DivideBy0 View Post
    When you talk about primitives, do you mean a given integral may have two answers which don't actually equal each other but are nonetheless valid evaluations of the integral?

    yes, they only differ by constants..
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  7. #7
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    Quote Originally Posted by ebonyscythe View Post
    Here's the basic question:

    We have an integral, that of which we do not know, where two out of three students got the correct answer. Which two students are right?

    a. sin^2x+C
    b. -cos^2x+C
    c. -sin^2x+C

    My logic is that one of the (-)sin^2x+C is right, meaning the integral might be \int2sin(x)cos(x), but I'm not sure how one could get a -cos^2x+C out of that.

    Any help is appreciated.
    Is it not the case that:

    \sin^2(x)=1-\cos^2(x)?

    ZB
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  8. #8
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Constatine11 View Post
    Is it not the case that:

    \sin^2(x)=1-\cos^2(x)?

    ZB
    it can be the case.. take note of choice B.. C takes charge the "1" in your right hand side..
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