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Stefana.
Pe: 2014-10-05T07:50:43+03:00 In: In:

Integrale ciudate.

1)\int \left ( 2sin^{2}\frac{x}{2}+4cos^{2}\frac{x}{2} \right )dx, x\in \mathbb{R}

2)\int arcsin(sin x)dx, x\in \left [ 0, 2\pi \right ]

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  1. ghioknt
    profesor

    \int arcsin(sin x)dx, \;x\in \left [ 0, 2\pi \right ].
    Tinem cont ca arcsin:[-1;\,1]\to \left [ -\frac{\pi }{2};\,\frac{\pi }{2} \right ] este inversa functiei \sin :\left [ -\frac{\pi }{2};\,\frac{\pi }{2} \right ]\to [-1;\,1].
    Asadar arcsin(sin x)=x,\;\forall x\in \left [ -\frac{\pi }{2};\,\frac{\pi }{2} \right ].
    Pentru\;x\in \left ( \frac{\pi }{2};\,\frac{3\pi }{2} \right ],\;x-\pi \in \left [ -\frac{\pi }{2};\,\frac{\pi }{2} \right ]; atunci
    arcsin(sinx)=arcsin(-sin (x-\pi))=-arcsin(sin (x-\pi))=-x+\pi.
    Pentru\;x\in \left(\frac{3\pi}{2};\,2\pi\right ],\;x-2\pi \in \left -\frac{\pi}{2};\,0\right ]\;si\;arcsin(sin x)=arcsin(sin (x-2\pi ))=x-2\pi.
    Asadar functia de integrat este
    f(x)=\begin{cases}x\;\;\;pt.\;x\in [0;\,\frac{\pi}{2}]\\-x+\pi\;pt.x\in (\frac{\pi}{2};\,\frac{3\pi}{2}]\\x-2\pi\;pt.\;x\in (\frac{3\pi}{2};\,2\pi]\end{cases}, pentru care o primitiva trebuie sa fie de forma:
    F(x)=\begin{cases}\frac{x^2}{2}\;\;\;\;pt.\;x\in [0;\,\frac{\pi}{2}]\\-\frac{x^2}{2}+\pi x+c_2\;pt.x\in (\frac{\pi}{2};\,\frac{3\pi}{2}]\\\frac{x^2}{2}-2\pi x+c_3\;pt.\;x\in (\frac{3\pi}{2};\,2\pi]\end{cases}; am luat c_1=0.
    Pentru continuitate in \frac{\pi}{2}:\;\frac{\pi^2}{8}=-\frac{\pi^2}{8}+\pi \frac{\pi}{2}+c_2\Rightarrow c_2=-\frac{\pi^2}{4}.
    Pentru continuitate in \frac{3\pi}{2}:\;-\frac{9\pi^2}{8}+\pi \frac{3\pi}{2}-\frac{\pi^2}{4}=\frac{9\pi^2}{8}-2\pi \frac{3\pi}{2}+c_3\Rightarrow c_3=2\pi^2.
    Utilizand Corolarul teoremei lui Lagrange se verifica F'(x)=f(x) si in cele doua puncte \frac{\pi}{2},\;\frac{3\pi}{2}.
    Astept si intrebari.

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