How do you evaluate the integral of arcsin(sin(x)) from 0 to 2pi?

[vDrag0n]

Homework Statement

int (1/(4-r^2)^0.5) dr dx, r=0 to 2sinx, x=0 to 2pi

Homework Equations

How to continue the integral of x

The Attempt at a Solution

I'm stuck at

int(arcsin(sinx)) dx, x=0 to 2pi

 

[Mark44]

Homework Statement

int (1/(4-r^2)^0.5) dr dx, r=0 to 2sinx, x=0 to 2pi

Homework Equations

How to continue the integral of x

The Attempt at a Solution

I'm stuck at

int(arcsin(sinx)) dx, x=0 to 2pi

This is your iterated integral:
$$\int_{x = 0}^{2\pi}\int_{r = 0}^{2 sin(x)}\frac{1}{\sqrt{4 - r^2}}dr~dx$$

I'm pretty sure you evaluated the inner integral incorrectly, most likely because you have a mistake in your trig substitution.

For this integral
$$\int_0^{2 sin(x)}\frac{1}{\sqrt{4 - r^2}}dr$$
I get 2 sin(x)

That makes the outer integral pretty simple.