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Math Help - Trig-sub problem

  1. #1
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    Trig-sub problem

    \int{\frac{(1-x^2)^{3/2}}{x^2}}

    After my Trig sub , i get \frac{1}{x}+2arccos(x)-\frac{x^3}{3}+C

    when i differentiate that, i do not get my initial integral which is a problem , can someone do this integral step by step please
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by Larrioto View Post
    \int{\frac{(1-x^2)^{3/2}}{x^2}}

    After my Trig sub , i get \frac{1}{x}+2arccos(x)-\frac{x^3}{3}+C

    when i differentiate that, i do not get my initial integral which is a problem , can someone do this integral step by step please

    Here is how I approached it
    x =cos(t)

    dx = - sin(t)dt


    -\int{\frac{(sin(t))^{3}\times sin(t) }{cos^2(t)}}

    -\int{\frac{(sin^{4}(t)) }{cos^2(t)}}

    -\int{\frac{(sin^{2}(t))(1-cos^2(t)) }{cos^2(t)}}


    -\int{tan^{2}(t)-sin^2(t) }

    - tan(t) + t + \int{\frac{1-cos(2t)}{2}}

    I wont go further before you try it
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  3. #3
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    -\int{\frac{(sin(t))^{3}\times sin(t) }{cos^2(t)}}

    this is false, because the exponant is not 3, but 3/2
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  4. #4
    Moo
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    Quote Originally Posted by Larrioto View Post
    -\int{\frac{(sin(t))^{3}\times sin(t) }{cos^2(t)}}

    this is false, because the exponant is not 3, but 3/2
    (1-\cos^2(x))^{3/2}=(\sin^2(x))^{3/2}=\sin(x)^{2 \times 3/2}=\dots
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  5. #5
    MHF Contributor matheagle's Avatar
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    The error is here.....


    Quote Originally Posted by ADARSH View Post
    Here is how I approached it
    x =cos(t)

    dx = - sin(t)dt


    -\int{\frac{(sin(t))^{3}\times sin(t) }{cos^2(t)}}

    -\int{\frac{(sin^{4}(t)) }{cos^2(t)}}

    -\int{\frac{(sin^{2}(t))(1-cos^2(t)) }{cos^2(t)}}


    -\int{tan^{2}(t) TIMES  sin^2(t) }

    - tan(t) + t + \int{\frac{1-cos(2t)}{2}}

    I wont go further before you try it
    It's times and not subtraction -\int{\frac{sin^{4}(t)}{cos^2(t)}} =-\int{\frac{sin^{2}(t) }{cos^2(t)}}\sin^2t

    maybe try the two half angle formulas here instead of going with tangent times sine
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  6. #6
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by ADARSH View Post

    Here is how I approached it
    x =cos(t)

    dx = - sin(t)dt


    -\int{\frac{(sin(t))^{3}\times sin(t) }{cos^2(t)}}

    -\int{\frac{(sin^{4}(t)) }{cos^2(t)}}

    -\int{\frac{(sin^{2}(t))(1-cos^2(t)) }{cos^2(t)}}

    -\int{\frac{(sin^{2}(t)\times(1) }{cos^2(t)}}-\frac{sin^2(t)\times cos^2(t)}{cos^2(t)}


    -\int{tan^{2}(t)-sin^2(t) }

    - tan(t) + t + \int{\frac{1-cos(2t)}{2}}

    I wont go further before you try it
    *I dont understand it
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  7. #7
    MHF Contributor matheagle's Avatar
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    It looks fine. I guess through trig the integrand can be either.
    I figured you had to have a mistake, but they both seem ok.
    Now you need to get back to the x.
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  8. #8
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by ADARSH View Post

    - tan(t) + t + \int{\frac{1-cos(2t)}{2}}
    -tan(t) + \frac{3t}{2}-sin(2t)/4

    cos(t)=x

    tan(t) = \sqrt{\frac{1}{cos^2(t)} - 1}=\sqrt{\frac{1}{x^2} - 1}

    t =cos^{-1}(x)

    sin(2t)/4 = \frac{ sin(t)cos(t)}{2} = \frac{\sqrt{(1-cos^2(t))}\times cos(t)}{2} = \frac{\sqrt{(1-x^2)}\times x}{2}


    Thus we get our answer
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