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Math Help - Integration

  1. #1
    Newbie masterjoint's Avatar
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    Integration

    there are 3 some questions from my Calculus homework that are giving me real head ache ! , please help dudes

    have to get the integral of the followings
    1. dx / ( 2e^x ) * sqr root ( 1 - e^2x ) INDEFINITE

    2. { sqr root [ (1 - x^2)^3 ] } dx [0,1] DEFINITE

    3. { x^3 * sin^2 * x / x^4 + 2x^2 + 1 } dx [-5,5] DEFINITE


    please write the way of how to do each question with formulas used, it would be very very nice of you ! thanks alot in advanced !
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Smile ...

    Hello & a very warm Welcome to the forum
    Quote Originally Posted by masterjoint View Post
    1. dx / ( 2e^x ) * sqr root ( 1 - e^2x )
     <br />
I=\int{\frac{dx}{2e^x * \sqrt{1-e^{2x}}}} <br />

    put e^x=t

    so dx=\frac{dt}{t}

    hence
     <br /> <br />
I=\int{\frac{dt}{2t^2 * (1-t^2)^{1/2}}}<br />

    so

    I=\int{\frac{dt}{2t^3 * \sqrt{(\frac{1}{t^2} - 1)}}}

    put \frac{1}{t^2}= k

    so
     <br /> <br />
-2t^{-3} dt = dk<br />

    hence
     <br /> <br />
I=\int { \frac{dk}{-4*\sqrt{k-1}}}<br />

    put (k-1)=g

    so
    I=\int { \frac{dg}{-4*\sqrt{g}}}

    I am too lazy to continue so go ahead if unsolved feel free to ask
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  3. #3
    Eater of Worlds
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    The second one is difficult to integrate by elementary methods. But, I do not think it is required. It is a matter of observation to note we have an odd function. Since we integrate from -5 to 5, it evaluates to 0.
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