how to integrate sin(x/y) dx

**john1985** · April 23rd 2010, 07:26 PM

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)

**dwsmith** · April 23rd 2010, 09:00 PM

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

**Pulock2009** · April 23rd 2010, 09:03 PM

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.

**john1985** · April 24th 2010, 10:23 PM

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 ???

**maddas** · April 24th 2010, 10:37 PM

Is y a constant? If so: let x = yt, so dx = y dt. Then $\int \sin (\frac xy)\;\mathrm{d}x = \int \sin (t) \cdot y\;\mathrm{d}t = y \int \sin (t) \;\mathrm{d}t = -y \cos t + c = -y\cos(\frac xy) + c$ .

**mr fantastic** · April 25th 2010, 04:47 AM

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.

**dwsmith** · April 25th 2010, 11:14 AM

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

Try doing dy now.

**mr fantastic** · April 25th 2010, 03:14 PM

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:

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.

**dwsmith** · April 25th 2010, 03:19 PM

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.

**maddas** · April 25th 2010, 03:25 PM

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...

**mr fantastic** · April 25th 2010, 03:44 PM

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.

**sr917** · April 25th 2010, 09:52 PM

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

**mr fantastic** · April 26th 2010, 12:10 AM

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).

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