Antiderivative of Cotangent and Arcsine - Explanation Welcome

[mcmw]

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

 

[d_leet]

Express cotangent in terms of sines and cosines and then make an appropraite substitution to find its antiderivative, use integration by parts for arcsine.

 

[Matthew Rodman]

You can always try the Integrator. (Making sure, of course, that you triple-check the answer works before doing anything with it!)

 

[x.users]

mcmw

mm you can visit this web to know

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

you are welcome .

 

[cristo]

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!

 

[Qyzren]

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

 

[VietDao29]

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

$$\int \cot x dx = \int \frac{\cos x}{\sin x} dx$$
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:
$$\int \cot x dx = \int \frac{\cos x}{\sin x} dx = \int \frac{du}{u} = \ln |u| + C = \ln |\sin x| + C$$

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The antiderivative of arcsin can be found by Integration by Parts:
$$\int \arcsin x dx$$
$$u = \arcsin x \Rightarrow du = \frac{dx}{\sqrt{1 - x ^ 2}}$$
dv = dx ~~~> v = x
So, your original integral will become:
$$\int \arcsin x dx = x \arcsin x - \int \frac{x dx}{\sqrt{1 - x ^ 2}} = x \arcsin x + \frac{1}{2} \int \frac{d \left( 1 - x ^ 2 \right)}{\sqrt{1 - x ^ 2}} = x \arcsin x + \sqrt{1 - x ^ 2} + C$$

 

[mcmw]

Thank you

Thank you for the help everyone I completed the problems.