How to evaluate this double integral?

Kuma
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Homework Statement



∫0 to 2 ∫x/4 to 1/2 (sin (pi*y2)) dy dx

Homework Equations





The Attempt at a Solution



I think I have to convert this to polar or do some sort of change of variable.

Although in polar y = r sin θ, so then you would have sin of a sin??
 
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Try drawing the region and switching the order of integration, and this might allow you to use substitution rule hopefully, as I believe you will get something of the form \int y*\sin(\pi y^2) dy
 
Kuma said:

Homework Statement



∫0 to 2 ∫x/4 to 1/2 (sin (pi*y2)) dy dx

Homework Equations





The Attempt at a Solution



I think I have to convert this to polar or do some sort of change of variable.

Although in polar y = r sin θ, so then you would have sin of a sin??

Sketch the region (in the xy-plane) over which the integration is being done. Use this to switch the order of integration.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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