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Thread: Integration

  1. #1
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    Integration

    Hi Everyone,
    Please help me with this integration:
    $\displaystyle \int_0^\pi {\sqrt {2 + 2\sin (x)} } dx$

    This is what I've tried so far:
    when I substitute
    $\displaystyle z = 1 + \sin (x)$
    and
    $\displaystyle x \to \pi ,z \to 1$
    $\displaystyle x \to 0,z \to 1$

    I got:
    $\displaystyle \sqrt 2\int_1^1 {\frac{1}{{\sqrt {2 - z} }}dz} $

    the new boundaries is my main problem here, it will give me a value of 0.
    However, when I check my answer, it should be
    $\displaystyle 4\sqrt 2 $
    Last edited by HaiLiang96; Oct 1st 2017 at 09:16 PM.
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  2. #2
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    Re: Integration

    The problem is that when you replace the limits, you're changing the definition for the path of integration. Usually just substituting the limits works: if $\displaystyle s=4t$, then "s goes from 4 to 12" and "t goes from 1 to 3" mean the same thing.

    In your case, however, since $\displaystyle z=1+\sin x$, as x goes from 0 to $\displaystyle \pi$, z goes from 1 to 2 (when $\displaystyle x=\frac{\pi}{2}$) and then back to 1 again. That's not necessarily a problem, but your equation for x is different for the two ranges.

    For x between 0 and $\displaystyle \frac{\pi}{2}$, we want $\displaystyle x=\sin^{-1}(z-1)$, but for x between $\displaystyle \frac{\pi}{2}$ and $\displaystyle \pi$, we want $\displaystyle x=\pi-\sin^{-1}(z-1)$ (since the inverse sine function is defined to be between $\displaystyle -\frac{\pi}{2}$ and $\displaystyle \frac{\pi}{2}$). When you plug into the integrand, you get the same thing, but when you plug into dx, you get a negative sign on one but not the other.

    Once you see the problem, you work around it: since the integrand is symmetric around $\displaystyle x=\frac{\pi}{2}$,
    $\displaystyle \int_0^\pi \sqrt{2-2\sin x}\,dx = 2\int_0^\frac{\pi}{2} \sqrt{2-2\sin x}\,dx = 2\sqrt{2}\int_1^2 \frac{1}{\sqrt{2-z}}\,dz=4\sqrt{2}$.

    - Hollywood
    Thanks from HaiLiang96 and topsquark
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  3. #3
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    Integration

    Hi Hollywood,
    Thanks for answering....
    Can I say:
    whenever the curve is oscillating along x-axis (or it is a one-to-many function) when we integrate it by substitution, we should break it into few parts (local minimum to local maximum / local maximum to local minimum) to solve it?
    Last edited by HaiLiang96; Oct 2nd 2017 at 04:43 PM.
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  4. #4
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    Re: Integration

    No, the substitution method can still work (and usually does work, in my experience) even if the substitution function is not one-to-one. For example,

    $\displaystyle \int_{-1}^2 2x^3 \,dx = \frac{2^4}{2} - \frac{(-1)^4}{2} = 8 - \frac{1}{2} = \frac{15}{2}$

    Substituting $\displaystyle u=x^2$, $\displaystyle du=2x\,dx$ gives

    $\displaystyle \int_1^4 u \,du = \frac{4^2}{2} - \frac{1^2}{2} = 8 - \frac{1}{2} = \frac{15}{2} $

    but as x goes from -1 to 2, u goes from 1 to 0 and then back to 1 on its way to 4.

    - Hollywood

    By the way, in my previous post the integrand should be $\displaystyle \sqrt{2+2\sin{x}}$, not $\displaystyle \sqrt{2-2\sin{x}}$. Sorry about that.
    Thanks from HaiLiang96
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  5. #5
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    Re: Integration

    Another point: If you set z= 1+ sin(x) then dz= cos(x) dx. You cannot just replace "dx" with "dz". That substitution simply will not work here.
    Thanks from HaiLiang96
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  6. #6
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    Re: Integration

    Thanks for reply...
    I'm understand better now


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