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Homework Help: Inverse trig functions and pythagorean identity

  1. Apr 20, 2010 #1
    Hi. I'm having trouble trying to understand the relationship between inverse trig functions, especially arcsin x, and pythagorean identity. I know that because cosx=sqrt(1-(sinx)^2), derivative of arcsin x is 1/(cos(arcsin x)) = 1/(sqrt(1-(sinx)^2)arcsinx)) = 1/(sqrt(1-x^2). But how does pythagorean identity relate with antiderivative of arcsin, which is x arcsin x + sqrt(1-x^2) + C?
     
  2. jcsd
  3. Apr 20, 2010 #2

    Mark44

    Staff: Mentor

    I don't see that the pythagorean identity enters into it at all. To find the antiderivative of arcsin(x), use integration by parts, with u = arcsin(x) and dv = dx. If you don't understand how this works, let me know and I'll fill in the details.
     
  4. Apr 21, 2010 #3
    Mark44: Actually, I would be thankful if you can explain how the antiderivative of arcsin (x) works, because I have trouble understanding the concept. Thanks.
     
  5. Apr 21, 2010 #4

    Mark44

    Staff: Mentor

    Let u = arcsin(x) and dv = dx
    So du = dx/sqrt(1 - x^2), v = x
    [tex]\int arcsin(x) dx = x*arcsin(x) - \int \frac{x dx}{\sqrt{1 - x^2}}[/tex]

    For the integral on the right, let w = 1 - x^2, so dw = -2xdx

    [tex]= x*arcsin(x) - (-1/2)\int u^{-1/2}du = x*arcsin(x) + u^{1/2} + C[/tex]
    [tex]= x*arcsin(x) + \sqrt{1 - x^2} + C[/tex]
     
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