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I'm not really sure at all how to integrate this function. I was wondering if anyone could help me out,

$$\int^4_0 \int^2_{\sqrt y} \cos(x^3) \, dx \, dy$$

The answer choices are ${1 \over 3}\sin(64);\ \sin(8);\ \cos(8)-1; \ {1 \over 3}\sin(8); \ {1 \over 3}\sin(2);\ \sin(2)$

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You should reverse the order of integration. Draw a picture of the integration region and rotate by 90 degrees. The result should be

$$\int_0^2 dx \, \int_0^{x^2} dy \, \cos{x^3} = \int_0^2 dx \, x^2 \, \cos{x^3} $$

This you should be able to do.

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  • $\begingroup$ 'rotate by 90 degrees' is a curious expression here. Do you just mean change order of integration? $\endgroup$ – Simon S Dec 17 '15 at 1:22
  • $\begingroup$ This might be out of context, but: in which area of science is the notation $\int_0^2dx\int_0^{x^2}dy\cos x^3$ used instead of $\int_0^2\int_0^{x^2}\cos x^3dydx$ ? I've never seen the former used in a real analysis book. $\endgroup$ – Guest Dec 17 '15 at 1:36
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    $\begingroup$ @Guest: I've lost count how many times I have answered this question here. My background is in optics, where we do loads of double integrals. I find this notation easier on the eyes and is more operator-like. It has become habit to me. $\endgroup$ – Ron Gordon Dec 17 '15 at 1:38
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The function $\cos x^3$ does not have an elementary antiderivative, so you won't be able to evaluate the double integral in the current order.

So, let's switch the order of integration. The bounds of the region are $0 \le y \le 4$ and $\sqrt{y} \le x \le 2$. This can be combined as $0 \le y \le x^2 \le 4$. So, the new bounds are $0 \le x \le 4$ and $0 \le y \le x^2$.

This gives us $\displaystyle\int_{0}^{4}\int_{\sqrt{y}}^{2}\cos x^3\,dx\,dy = \int_{0}^{4}\int_{0}^{x^2}\cos x^3\,dy\,dx$.

The inner integral is easy to evaluate since the integrand is constant with respect to $y$. Once you evaluate the inner integral, the outer integral can be handled with a simple substitution.

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