# Antiderivative help?

## Main Question or Discussion Point

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

## Answers and Replies

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

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

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

Last edited:
VietDao29
Homework Helper

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:
$$\int \arcsin x dx = x \arcsin x - \int \frac{x dx}{\sqrt{1 - x ^ 2}} = x \arcsin x + \frac{1}{2} \int \frac{d \left( 1 - x ^ 2 \right)}{\sqrt{1 - x ^ 2}} = x \arcsin x + \sqrt{1 - x ^ 2} + C$$

Thank you

Thank you for the help everyone I completed the problems.