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  1. MHB Apprentice

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    #1
    So I have to show 2sin^2(x) and -cos(2x) have the same antiderative.

    Here's how I approached this.

    2sin^2(x) = 1-cos2x ==> u = 2x
    intergral of that is
    (u - sinu)/2 + c = x - (sinx)/2 + c

    -cos2x ==> u = 2x
    intregal of that is
    (-sinu)/2 + c= -(sin2x)/2 + c

    Have I calculated/approached this exercise wrongly? I don't see how they could be the same antiderative
    I have another similar exercise with 2cos^2(x) and cos(2x).

  2. Pessimist Singularitarian
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    #2
    Since this question involves integral calculus, I have moved it here.

    In order for two functions to have the same anti-derivative, they must in fact be equivalent. From a double-angle identity for cosine, we know:

    $ \displaystyle \cos(2x)=1-2\sin^2(x)$

    Hence:

    $ \displaystyle -\cos(2x)=2\sin^2(x)-1\ne2\sin^2(x)$

    Thus, the two given functions are not equivalent, therefore they cannot have the same anti-derivative.

    Can you use a similar line of reasoning for the second question?

  3. MHB Apprentice

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    #3 Thread Author
    Thanks alot for the confirmation, way the question was presented was very confusing to me, as if they had to have the same antideravative

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    #4
    I think the question actually must be to show that the two functions have the same derivative.

    As $\displaystyle \begin{align*} -\cos{(2\,x)} \equiv -\left[ 1 - 2\sin^2{(x)} \right] \equiv 2\sin^2{(x)} -1 \end{align*}$, which only differs from $\displaystyle \begin{align*} 2\sin^2{(x)} \end{align*}$ by a constant, they will in fact have the same derivatives.

  5. MHB Apprentice

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    #5 Thread Author
    This must be the answer I needed, thank you

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