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

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- Thread starter mcmw
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- #1

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

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mcmw

mm you can visit this web to know

http://en.wikipedia.org/wiki/Antiderivative

you are welcome .

mm you can visit this web to know

http://en.wikipedia.org/wiki/Antiderivative

you are welcome .

- #5

cristo

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mcmw

mm you can visit this web to know

http://en.wikipedia.org/wiki/Antiderivative

you are welcome .

I don't see the antiderivative of the arcsine, or cotangent functions on that webpage!

- #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

arcsin x = d/dx [(1-x²)^½ + x*arcsin x] <--Use integration by parts

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- #7

VietDao29

Homework Helper

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[tex]\int \cot x dx = \int \frac{\cos x}{\sin x} dx[/tex]

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

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

[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]

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Thank you for the help everyone I completed the problems.

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