# how to integrate sin(x/y) dx

• April 23rd 2010, 08:26 PM
john1985
how 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)
• April 23rd 2010, 10:00 PM
dwsmith
Quote:

Originally Posted by john1985
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)

When you integrate with respect to x, treat y as a constant.
When you integrate with respect to y, treat x as a constant
• April 23rd 2010, 10:03 PM
Pulock2009
suggestion
Quote:

Originally Posted by john1985
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)

t=x/y.
• April 24th 2010, 11:23 PM
john1985
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 ???
• April 24th 2010, 11:37 PM
Is y a constant? If so: let x = yt, so dx = y dt. Then $\int \sin (\frac xy)\;\mathrm{d}x = \int \sin (t) \cdot y\;\mathrm{d}t = y \int \sin (t) \;\mathrm{d}t = -y \cos t + c = -y\cos(\frac xy) + c$.
• April 25th 2010, 05:47 AM
mr fantastic
Quote:

Originally Posted by john1985
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)

You might not think that the region of integration is important, but it is. Post the whole question please if you hope to get effective help.
• April 25th 2010, 12:14 PM
dwsmith
$\int sin \bigg( \frac{x}{y} \bigg)dx=y\int sin \bigg(\frac{x}{y} \bigg)\bigg(\frac{1}{y}dx\bigg)=-ycos\bigg(\frac{x}{y}\bigg)+C$

Try doing dy now.
• April 25th 2010, 04:14 PM
mr fantastic
Quote:

Originally Posted by dwsmith
$\int sin \bigg( \frac{x}{y} \bigg)dx=y\int sin \bigg(\frac{x}{y} \bigg)\bigg(\frac{1}{y}dx\bigg)=-ycos\bigg(\frac{x}{y}\bigg)+C$

Try doing dy now.

I doubt this is what the question has asked the OP to do. See the boldface in the quote below:

Quote:

Originally Posted by john1985
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)

Until what I've boldfaced is explained, the real question cannot be reliably answered.
• April 25th 2010, 04:19 PM
dwsmith
Quote:

Originally Posted by mr fantastic
I doubt this is what the question has asked the OP to do. See the boldface in the quote below:

Until what I've boldfaced is explained, the real question cannot be reliably answered.

I am under the impression that he wants to integrate with dx and then a separate integral of dy.
• April 25th 2010, 04:25 PM
Quote:

Originally Posted by dwsmith
I am under the impression that he wants to integrate with dx and then a separate integral of dy.

Doubtful, since integrating again w.r.t. y will require non-elementary functions...
• April 25th 2010, 04:44 PM
mr fantastic
Quote:

Originally Posted by dwsmith
I am under the impression that he wants to integrate with dx and then a separate integral of dy.

I'll bet dollars to doughnuts that the original question gives a region that has to be integrated over. But we will never know unless the OP replies. Further posts are useless until then.
• April 25th 2010, 10:52 PM
sr917
evaluate the double integral sin(x/y)dA where R is the region bounded by the y-axis , y=pi and x=y^2
• April 26th 2010, 01:10 AM
mr fantastic
Quote:

Originally Posted by sr917
evaluate the double integral sin(x/y)dA where R is the region bounded by the y-axis , y=pi and x=y^2

Draw the region of integration. It is then clear that the required integral is

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

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

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

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