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)
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 ???
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).