[SOLVED] Need Help With a Double Integral

[Ithryndil]

Any help on the following integral would be appreciated. I don't know where to begin at all.


\int_0^{1}\int_{arcsiny}^{\pi/2} sec^2(cosx) dxdy


I've thought about changing the order of integration, but I don't think that will help. I am not asking someone to solve this at all, just to point me in the right direction. Thanks for any and all help.

[rocomath]

I haven't put it on paper but looks crazy.

Try changing it to cosine and then using double angle identity.

[Ithryndil]

I don't even know where to begin. Yes the integral of sec^2(x) is tan(x)....but you can't do that.

[tiny-tim]

I've thought about changing the order of integration, but I don't think that will help.

Hi Ithryndil! :smile:

erm … don't think about changing the order of integration … do it! :rolleyes:

You'll find it works! :smile:

[Ithryndil]

Well, I did it and do you have to write to different double integrals to describe the region?

[tiny-tim]

Hi Ithryndil! :smile:

(btw, it would help you if actually showed us what you've tried)

The chief problem in changing the order of integration is that you have to change the limits of the two integrals.

In this case, the limits are arcsiny < x < π/2 and 0 < y < 1.

You have to change this to the form f(x) < y < g(x) and a < x < b, where a and b are constants.

Hint: to find f and g, you must combine arcsiny < x with 0 < y < 1. :smile:

[Ithryndil]

Yea, I know it would help but I had trouble last night with the coding...anyhow, when changing the order of integration I got the following which should be correct.



\int_0^{\pi/2}\int_{0}^{sinx} sec^2(cosx) dydx





\int_0^{\pi/2} ysec^2(cosx) dx



The integral goes from 0 to sinx and you wind up getting



\int_0^{\pi/2} sinxsec^2(cosx) dx



Then you let u = cosx and get du = -sinxdx. From there it's a u substitution. At least, that's what I believe...Like I said, last night I was having issues with the formatting not working out for the forums and so I didn't want to try and show all of my work.

Below is the graph of the region (Or should be provided I understand this right).

[tiny-tim]

:biggrin: Woohoo! :biggrin:

Yes … as you see, it's just a question of changing the limits (and not panicking)!

(I assume you meant \int_0^{\pi/2} \left[ysec^2(cosx)\right]_0^{sinx} dx, but couldn't do the latex?)

If you're happy, try putting x = θ (an angle), and plotting the area with coordinates y and θ: what do you get? :smile:

[Ithryndil]

Yea, I couldn't do the latex...so I just put the little message below it.

[tiny-tim]

Yea, I couldn't do the latex...so I just put the little message below it.

Hi Ithryndil! :smile:

Yes, I thought so … if you click on QUOTE under my post, you'll be able to copy the code for it.

(And you may like to bookmark http://www.physics.udel.edu/~dubois/lshort2e/node61.html#SECTION008100000000000000000
and maybe http://www.physics.udel.edu/~dubois/lshort2e/node54.html#SECTION00830000000000000000
for more tips.)

Did you try the y,θ diagram? :smile: