# Thread:

1. ## 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)

2. 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

3. ## suggestion

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.

4. 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 ???

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

6. 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.

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

Try doing dy now.

8. 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.

9. 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.

10. 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...

11. 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.

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

13. 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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