Solving a double integral

  • Thread starter vDrag0n
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vDrag0n
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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
 

Answers and Replies

  • #2
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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:
[tex]\int_{x = 0}^{2\pi}\int_{r = 0}^{2 sin(x)}\frac{1}{\sqrt{4 - r^2}}dr~dx[/tex]

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

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

That makes the outer integral pretty simple.
 

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