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Thread: cos(SQRT{x}) dx (some work included).

  1. #1
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    cos(SQRT{x}) dx (some work included).

    Hello guys here is my problem uploaded: , see the attached image of my scanned paper

    The thing that i wonder about here is the hint : 2$cos u * u * dx-

    Where this this expression come from, how did they apply Cos to .where did it suddenly come from. I can understand why the constant was suddenly omitted from the right side of the equation (Constant Multiplie Rule). Would be happy if someone could solve this for me.

    cos(SQRT{x}) dx (some work included).-img_0001-001.jpg
    Last edited by mr fantastic; Jan 15th 2011 at 12:21 PM. Reason: Title.
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  2. #2
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    Let me just retype what you have so others can read it easier:

    $\displaystyle \displaystyle \int \cos \sqrt x\, dx$

    Substitue $\displaystyle x = u^2$
    $\displaystyle dx = 2u\,du$

    $\displaystyle \displaystyle \int \cos \sqrt x\, dx = \int \cos u\, 2u\, du = 2\int u\cos u\, du$

    Is this where you are stuck?

    Try to do integration by parts.
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  3. #3
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    We won't solve it for you, but the trick is to try to get the derivative of $\displaystyle \displaystyle \sqrt{x}$, i.e. $\displaystyle \displaystyle \frac{1}{2\sqrt{x}}$, as a factor.

    So $\displaystyle \displaystyle \int{\cos{\sqrt{x}}\,dx} = \int{\frac{2\sqrt{x}\cos{\sqrt{x}}}{2\sqrt{x}}\,dx }$

    $\displaystyle \displaystyle = 2\int{\sqrt{x}\cos{(\sqrt{x})}\,\frac{1}{2\sqrt{x} }\,dx}$.

    Now make the substitution $\displaystyle \displaystyle u = \sqrt{x}$ so that $\displaystyle \displaystyle \frac{du}{dx}= \frac{1}{2\sqrt{x}}$ and the integral becomes

    $\displaystyle \displaystyle 2\int{u\cos{u}\,\frac{du}{dx}\,dx}$

    $\displaystyle \displaystyle = 2\int{u\cos{u}\,du}$.


    You can now solve this using integration by parts.
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  4. #4
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    Snowteas and prove its ways of solving are different, Snowtea you based your on my variable substitution by using u^2.

    Prove it could you elaborate on your expression of getting to: \displaystyle \int{\cos{\sqrt{x}}\,dx} = \int{\frac{2\sqrt{x}\cos{\sqrt{x}}}{2\sqrt{x}}\,dx } because all I can understand now after looking at the derivate of sqrt[x] is that the derivate is given by using "Difference of two squares" formula and then you get 1/2*sqrt[x]. Thats understandable after trying it out myself half an hour ago.

    Also both of your answers lead to the same integral, here is where i am stuck again.


    I have integrated by using the DI shorcut (http://www.delmar.edu/math/MLC/Forms...By%20Parts.pdf) for fast results now.


    Here is my solution Tell me if it's appropriate or not.
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  5. #5
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    Riazy, your solution is really hard to read.
    Can you try to retype it in latex?

    For example typing
    [tex]\displaystyle \int \frac{x^2}{\cos(\sqrt{x})} dx[/tex]
    gives
    $\displaystyle \displaystyle \int \frac{x^2}{\cos(\sqrt{x})} dx$.
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  6. #6
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    What Prove It did, was simply multiply by 1.

    $\displaystyle \displaystyle \int{\cos{\sqrt{x}}\,dx} = \displaystyle \int{\cos{\sqrt{x}} \times 1 \,dx} = \displaystyle \int{\cos{\sqrt{x}} \times \dfrac{2\sqrt{x}}{2\sqrt{x}}\,dx} = \int{\frac{2\sqrt{x}\cos{\sqrt{x}}}{2\sqrt{x}}\,dx }$

    EDIT:

    And no, that's not it. Integration by parts:

    $\displaystyle \displaystyle \int uv' dx = [uv] - \int vu' \ dx$

    Hence;

    $\displaystyle \displaystyle \int u\cos u\ du = [u \sin u] - \int \sin u \cdot 1\ du$

    Can you complete it now?
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