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Math Help - integral of inverse hyperbolic

  1. #1
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    integral of inverse hyperbolic

    I have to find the integral of (x^2)(arccothx) dx

    I have used integration by parts with
    u = arccothx
    du = 1/(1-x^2)
    dv = x^2
    v = 1/3*x^3

    which i get:

    (1/3)(x^3 * arccothx) - integral (x^3)/3(1-x^2)

    i am stuck here as I do not know if I have to use polynomial long division on the integrand above or what the next step might be.
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  2. #2
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    Quote Originally Posted by shadow85 View Post
    I have to find the integral of (x^2)(arccothx) dx

    I have used integration by parts with
    u = arccothx
    du = 1/(1-x^2)
    dv = x^2
    v = 1/3*x^3

    which i get:

    (1/3)(x^3 * arccothx) - integral (x^3)/3(1-x^2)

    i am stuck here as I do not know if I have to use polynomial long division on the integrand above or what the next step might be.
    Yes, use polynomial division to reduce the degree of the numerator:

    \dfrac{x^3}{3(x^2-1)} = -\frac13\,\dfrac{x^3}{x^2-1} = -\frac13\Bigl(x +\dfrac x{x^2-1}\Bigr).
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  3. #3
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    Now try a trigonometric substitution x = \sin{\theta} so that \theta = \arcsin{\theta} and dx = \cos{\theta}\,d\theta and the integral becomes

    \frac{1}{3}\int{\frac{x^3}{1 - x^2}\,dx} = \frac{1}{3}\int{\frac{\sin^3{\theta}}{1 - \sin^2{\theta}}\,\cos{\theta}\,d\theta}

     = \frac{1}{3}\int{\frac{\sin^3{\theta}\cos{\theta}}{  \cos^2{\theta}}\,d\theta}

     = \frac{1}{3}\int{\frac{\sin^3{\theta}}{\cos{\theta}  }\,d\theta}

     = \frac{1}{3}\int{\frac{\sin{\theta}(1 - \cos^2{\theta})}{\cos{\theta}}\,d\theta}

     = \frac{1}{3}\int{\frac{\sin{\theta} - \sin{\theta}\cos^2{\theta}}{\cos{\theta}}\,d\theta  }

     = \frac{1}{3}\left(\int{\frac{\sin{\theta}}{\cos{\th  eta}}\,d\theta} + \int{-\sin{\theta}\cos{\theta}\,d\theta}\right)

    Each of these is solvable using the substitution u = \cos{\theta}.
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    thank you guys.
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