Hi all,

I'm wondering if someone can point me in the right direction for integrating this integral...

$\displaystyle \int \sin{(\frac{x}{y})} dy$

I'm not sure where to start,

thanks all.

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- Apr 26th 2010, 11:39 PMPeleusIntegration help needed, thanks
Hi all,

I'm wondering if someone can point me in the right direction for integrating this integral...

$\displaystyle \int \sin{(\frac{x}{y})} dy$

I'm not sure where to start,

thanks all. - Apr 26th 2010, 11:46 PMmaddas
Let y = xu. Then it becomes $\displaystyle x\int \sin(\frac1u)du$ which is known not to have an antiderivative amoung the elementary functions. Wolfram Alpha can give it to you with special functions though.

- Apr 27th 2010, 12:41 AMPeleus
If anyone could check through my working it would be great.

Q. Evaluate the following double integral

$\displaystyle \int\int_R \sin{\frac{x}{y}} dA $

Where R is the region bounded by the y-axis, $\displaystyle y = \pi$, $\displaystyle x = y^2$

Drawing a horizontal line throughout the region, we end up with...

$\displaystyle \int^{\pi}_0 \int^{y^2}_0 \sin{\frac{x}{y}} dxdy $

This turns through integration into

$\displaystyle \int^{\pi}_0 [ -y\cos{\frac{x}{y}}]^{y^2}_0 dy $

Subbing in values turns it into

$\displaystyle \int^{\pi}_0 [-y\cos{y} + y] dy $

Integration by parts turning into

$\displaystyle [\frac{y^2}{2} - \cos{y} - y\sin{u}]^{\pi}_0 $

Giving

$\displaystyle \frac{\pi^2}{2} - \cos{\pi} - \pi\sin{\pi} + \cos{0} $

Am I right so far?

Thanks.