Antiderivative help?

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does anyone know the antiderivative of cotangent or of arcsine? any explanation would be appreciated.
 

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  • #2
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Express cotangent in terms of sines and cosines and then make an appropraite substitution to find its antiderivative, use integration by parts for arcsine.
 
  • #3
You can always try the Integrator. (Making sure, of course, that you triple-check the answer works before doing anything with it!)
 
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  • #5
cristo
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  • #6
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cot x = d/dx [ln (sin x)]
arcsin x = d/dx [(1-x²)^½ + x*arcsin x] <--Use integration by parts
 
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  • #7
VietDao29
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does anyone know the antiderivative of cotangent or of arcsine? any explanation would be appreciated.
[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]

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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]
 
  • #8
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Thank you

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

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