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Antiderivative help?

  1. Mar 15, 2007 #1

    does anyone know the antiderivative of cotangent or of arcsine? any explanation would be appreciated.
     
  2. jcsd
  3. Mar 15, 2007 #2
    Express cotangent in terms of sines and cosines and then make an appropraite substitution to find its antiderivative, use integration by parts for arcsine.
     
  4. Mar 16, 2007 #3
    You can always try the Integrator. (Making sure, of course, that you triple-check the answer works before doing anything with it!)
     
  5. Mar 16, 2007 #4
  6. Mar 16, 2007 #5

    cristo

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    I don't see the antiderivative of the arcsine, or cotangent functions on that webpage!
     
  7. Mar 21, 2007 #6
    cot x = d/dx [ln (sin x)]
    arcsin x = d/dx [(1-x²)^½ + x*arcsin x] <--Use integration by parts
     
    Last edited: Mar 21, 2007
  8. Mar 21, 2007 #7

    VietDao29

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    Homework Helper

    [tex]\int \cot x dx = \int \frac{\cos x}{\sin x} dx[/tex]
    Since, the power of cosine function is odd, we can let u = sin x.
    (In fact, the power of sine function is also odd, so letting u = cos x should be fine as well)
    u = sin x ~~~> du = cos x dx
    So, the integral becomes:
    [tex]\int \cot x dx = \int \frac{\cos x}{\sin x} dx = \int \frac{du}{u} = \ln |u| + C = \ln |\sin x| + C[/tex]

    ----------------
    The antiderivative of arcsin can be found by Integration by Parts:
    [tex]\int \arcsin x dx[/tex]
    [tex]u = \arcsin x \Rightarrow du = \frac{dx}{\sqrt{1 - x ^ 2}}[/tex]
    dv = dx ~~~> v = x
    So, your original integral will become:
    [tex]\int \arcsin x dx = x \arcsin x - \int \frac{x dx}{\sqrt{1 - x ^ 2}} = x \arcsin x + \frac{1}{2} \int \frac{d \left( 1 - x ^ 2 \right)}{\sqrt{1 - x ^ 2}} = x \arcsin x + \sqrt{1 - x ^ 2} + C[/tex]
     
  9. Mar 25, 2007 #8
    Thank you

    Thank you for the help everyone I completed the problems. :smile:
     
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