can anyone assist with evaluating the integral if sin (x/y) with respect to x & y, i have tried using integration by parts but cant seem to get an answer.

i know by using the online calculators the answer is -ycos(x/y)

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- Apr 23rd 2010, 07:26 PMjohn1985how to integrate sin(x/y) dx
can anyone assist with evaluating the integral if sin (x/y) with respect to x & y, i have tried using integration by parts but cant seem to get an answer.

i know by using the online calculators the answer is -ycos(x/y) - Apr 23rd 2010, 09:00 PMdwsmith
- Apr 23rd 2010, 09:03 PMPulock2009suggestion
- Apr 24th 2010, 10:23 PMjohn1985
would some be able to post some of the initial steps i get you have to use the substitution method but am having difficultiy getting the answer out

original expression integrate with respect to x

= sin(y/x)

step 2 using substitution = sin (t) dt

step 3 -cos (t) t ??? - Apr 24th 2010, 10:37 PMmaddas
Is y a constant? If so: let x = yt, so dx = y dt. Then $\displaystyle \int \sin (\frac xy)\;\mathrm{d}x = \int \sin (t) \cdot y\;\mathrm{d}t = y \int \sin (t) \;\mathrm{d}t = -y \cos t + c = -y\cos(\frac xy) + c$.

- Apr 25th 2010, 04:47 AMmr fantastic
- Apr 25th 2010, 11:14 AMdwsmith
$\displaystyle \int sin \bigg( \frac{x}{y} \bigg)dx=y\int sin \bigg(\frac{x}{y} \bigg)\bigg(\frac{1}{y}dx\bigg)=-ycos\bigg(\frac{x}{y}\bigg)+C$

Try doing dy now. - Apr 25th 2010, 03:14 PMmr fantastic
- Apr 25th 2010, 03:19 PMdwsmith
- Apr 25th 2010, 03:25 PMmaddas
- Apr 25th 2010, 03:44 PMmr fantastic
- Apr 25th 2010, 09:52 PMsr917
evaluate the double integral sin(x/y)dA where R is the region bounded by the y-axis , y=pi and x=y^2

- Apr 26th 2010, 12:10 AMmr fantastic
Draw the region of integration. It is then clear that the required integral is

$\displaystyle \int_{y = 0}^{y = \pi} \int_{x = 0}^{x = y^2} \sin \left(\frac{x}{y}\right) \, dx \, dy$

$\displaystyle = \int_0^\pi \left[-y \cos \left( \frac{x}{y}\right) \right]_0^{y^2} \, dy$

$\displaystyle = - \int_0^\pi y \cos (y) \, dy = ....$

(If only the complete question had been asked in the first place a lot of time and energy would not have been wasted).