Is It Possible to Simplify This Tricky Triple Integral?

[Kreamer]

[tex]\int[/tex][tex]^{1}_{0}[/tex][tex]\int[/tex][tex]^{x/2}_{0}[/tex][tex]\frac{y}{(2y-1)\sqrt{1+y^2}}[/tex]dydx

Most of my attempts at this problem fail pretty quickly. Not even my calculator knows what to do with this one.

 

[I like Serena]

[tex]\int^{1}_{0}\int^{x/2}_{0}\frac{y}{(2y-1)\sqrt{1+y^2}} dydx[/tex]

Most of my attempts at this problem fail pretty quickly. Not even my calculator knows what to do with this one.

This looks like another integral I just saw here.
I presume that one is yours as well?

This one looks equally difficult to solve.
My approach would be to approximate it numerically.
Is it possible that is intended?

It still means you need to bring in a suitable form to integrate numerically.
I guess you could then integrate it with your calculator?

The trick would be to swap the integrals, implying a change in boundary values.
Then you can easily integrate the inner integral over x, leaving the outer integral that will now have fixed boundaries.
Your calculator should be able to do the rest.

 

[SteamKing]

It's a double integral and it evaluates to 1 - SQRT (5) / 2

 

[SVXX]

Would a change in the order of integration help? :)

 

[SteamKing]

If you want to do the integration, I think you will have to look at integration by parts with respect to y, using either trigonometric or hyperbolic substitution for the function of y under the square root.

 

[grey_earl]

In the inner integral:
You could try substituting y = sinh z to make the square root go away. Then you split of a summand 1/2 which integrates to z/2, and to find an indefinite integral for the remaining part, you should calculate the derivative of artanh(a + b tanh(z/2)).

 

[SteamKing]

Here's my calculation, using Derive for Windows

 

[I like Serena]

The area that is integrated is a triangle between (0, 0), (1, 0), and (1, 1/2).
If we swap the integration order then y must run from 0 to 1/2, and the corresponding x must run from 2y to 1.

So we have:

[tex]\int^{1}_{0}\int^{x/2}_{0}\frac{y}{(2y-1)\sqrt{1+y^2}} dydx[/tex]

[tex]= \int^{1/2}_{0}\int^{2y}_{1}\frac{y}{(2y-1)\sqrt{1+y^2}} dxdy[/tex]

[tex]= \int^{1/2}_{0}\frac{y}{(2y-1)\sqrt{1+y^2}} (1 - 2y)dy[/tex]

[tex]= \int^{1/2}_{0}\frac{-y}{\sqrt{1+y^2}}dy[/tex]

[tex]= \left.-\sqrt{1+y^2}}\right|_{0}^{1/2}[/tex]

[tex]= 1 - \frac 1 2 \sqrt{5}[/tex]

It turns out to be easier than I thought .