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Thread: Trig-sub problem

  1. #1
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    Trig-sub problem

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

    After my Trig sub , i get $\displaystyle \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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    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by Larrioto View Post
    $\displaystyle \int{\frac{(1-x^2)^{3/2}}{x^2}}$

    After my Trig sub , i get $\displaystyle \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


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

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

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


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

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

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

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

    this is false, because the exponant is not 3, but 3/2
    $\displaystyle (1-\cos^2(x))^{3/2}=(\sin^2(x))^{3/2}=\sin(x)^{2 \times 3/2}=\dots$
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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


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

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

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


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

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

    I wont go further before you try it
    It's times and not subtraction $\displaystyle -\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


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

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

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

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


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

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

    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

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

    $\displaystyle cos(t)=x$

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

    $\displaystyle t =cos^{-1}(x)$

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


    Thus we get our answer
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