how to solve $\int_{0}^{4}\int_{\sqrt{x}}^{2} \sin(xy^3) dydx$?

Asked Dec 07 '20 at 15:21

Active Dec 08 '20 at 07:55

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$$\int_{0}^{4}\int_{\sqrt{x}}^{2} \sin(xy^3) dydx$$

My solution is: (I'm not sure if it's right.) $$\int_{0}^{4}\int_{\sqrt{x}}^{2} \sin(xy^3) dydx=\int_{0}^{2}\int_{0}^{y^2} \sin(xy^3) dxdy=-\int_{0}^{2}\frac{1-\cos(y^5)}{y^3}dy$$ but I don't know what to do with rest.

edited Dec 08 '20 at 07:55

asked Dec 07 '20 at 15:21

Elsa Hejazian

Your rightmost expression is wrong. It should be: $\displaystyle \cdots = \int \limits_0^2 \frac{1-\cos(y^5)}y , dy $
And I doubt there's a clean form to this integral

And yes, your middle double integral is correct.
– GohP.iHan Dec 07 '20 at 19:51
@GohP.iHan shouldn't the denominator of $\frac{1-\cos(y^5)}{y}$ be $y^3$? – Alann Rosas Dec 08 '20 at 08:34
Yes, my $\LaTeX$ skill failed me. I tried to fix my comment above, but I guess I can't edit it anymore after a couple of hours later. – GohP.iHan Dec 08 '20 at 08:36

how to solve $\int_{0}^{4}\int_{\sqrt{x}}^{2} \sin(xy^3) dydx$?

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