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Math Help - another integration - need help!

  1. #1
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    Question another integration - need help!

    8 times the integral ( from -1/sqrt(2) to 1/sqrt(2) ) of ( 1-2x^2)^3/2 dx

    Please help!! Thank you very much.
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  2. #2
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    A trig. substitution would work... even is not a cool integral...
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  3. #3
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    \int\limits_{ - \frac{1}{{\sqrt 2 }}}^{\frac{1}{{\sqrt 2 }}} {\sqrt[3]{{\left( {1 - 2x^2 } \right)^2 }}}  = 2\int\limits_0^{\frac{1}{{\sqrt 2 }}} {\sqrt[3]{{\left( {1 - 2x^2 } \right)^2 }}} <br />
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  4. #4
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    Hi plato,

    And then...

    Could you please show me the full steps ? Thank you very much.


    Hi Krizalid,

    Please show me the full steps with trig substitution. I still can't work to the end with it. I really forgot the trig substitution work .

    The previous question. I worked it out simply by using the face that the integral is equal to 1/2 pi a^2 .
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  5. #5
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    Hello, kittycat!


    8\int^{\frac{1}{\sqrt{2}}}_{-\frac{1}{\sqrt{2}}}\left(1-2x^2\right)^{\frac{3}{2}}\,dx

    Let: 2x^2 \:=\:\sin^2\!\theta\quad\Rightarrow\quad x \:=\:\frac{1}{\sqrt{2}}\sin\theta\quad\Rightarrow\  quad dx \:=\:\frac{1}{\sqrt{2}}\cos\theta\,d\theta
    . . and: \left(1 - 2x^2\right)^{\frac{3}{2}} \:=\:\left(1-\sin^2\!\theta\right)^{\frac{3}{2}} \:=\:\left(\cos^2\!\theta\right)^{\frac{3}{2}} \:=\:\cos^3\!\theta


    Substitute: . 8\int \cos^3\!\theta\left(\frac{1}{\sqrt{2}}\cos\theta\,  d\theta\right) \:=\:4\sqrt{2}\int\cos^4\!\theta\,d\theta \;=\;4\sqrt{2}\int\left(\cos^2\!\theta\right)^2d\t  heta

    . . = \;4\sqrt{2}\int\left(\frac{1 + \cos2\theta}{2}\right)^2d\theta \;=\;\sqrt{2}\int\left(1 + 2\cos2\theta + \cos^2\!2\theta\right)\,d\theta

    . . = \;\sqrt{2}\int\left(1 + 2\cos2\theta + \frac{1 + \cos4\theta}{2}\right)d\theta \;= \;\sqrt{2}\int\left(\frac{3}{2} + 2\cos2\theta + \frac{1}{2}\cos4\theta\right)d\theta


    Can you finish it now?

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  6. #6
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    As you can see kittycat, Soroban gave you a full trig. sub. that I was talkin' about
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  7. #7
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    Hi soroban,

    Thank you very much for your detailed work.

    I followed it but can't obtained the answer 3pi/2 ( the answer given by my teacher) . Do you get the final answer 3pi/2?
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  8. #8
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    Hello, kittycat!

    I don't get that answer either . . .


    We have: . \sqrt{2}\int \left(\frac{3}{2} + 2\cos2\theta + \frac{1}{2}\cos4\theta\right)\,d\theta \;= \;\sqrt{2}\left(\frac{3}{2}\theta + \sin2\theta + \frac{1}{8}\sin4\theta\right)

    . . . . . = \;\sqrt{2}\left[\frac{3}{2}\theta + 2\sin\theta\cos\theta + \frac{1}{4}\sin\theta\cos\theta(1 - 2\sin^2\theta)\right]

    Back-substitute: . \sqrt{2}\left[\frac{3}{2}\arcsin(\sqrt{2}x) + 2\sqrt{2}x\sqrt{1-2x^2} + \frac{1}{4}\sqrt{2}x\sqrt{1-2x^2}(1-4x^2)\right]^{\frac{1}{\sqrt{2}}} _{\text{-}\frac{1}{\sqrt{2}}}

    Evaluate: . \sqrt{2}\left[\left(\frac{3}{2}\arcsin(1) + 2\!\cdot\!0 + \frac{1}{4}\!\cdot\!0\right) - \left(\frac{3}{2}\arcsin(-1) + 2\!\cdot\!0 + \frac{1}{4}\!\cdot\!0\right)\right]

    . . . . . = \;\sqrt{2}\left[\frac{3}{2}\!\cdot\!\frac{\pi}{2} - \frac{3}{2}\left(\text{-}\frac{\pi}{2}\right)\right] \;=\;\sqrt{2}\left[\frac{3\pi}{4} + \frac{3\pi}{4}\right] \;=\;\sqrt{2}\!\cdot\!\frac{3\pi}{2}


    So I get: . \frac{3\sqrt{2}\pi}{2}\; \text{ or }\;\frac{3\pi}{\sqrt{2}}

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  9. #9
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    Hi soroban,

    May I ask you one more question for this question? Why let 2x^2 =sin^2 theta ? Thank you very much.

    I know in trig substitution, when you see sqrt ( a^2 -x^2) , you will let x = asintheta.
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