# Simplifying help !

1. Dec 3, 2005

### elle

Simplifying help plz!

Hi, I was given a double integral question and I managed to do the x integration. After placing the limits I get the following:

∫{ (2y²)(√2+y²) - (2y²)(√2y²) } dy

I know the integrand can be simplified but I don't have a clue. Can anyone help? Thank you!!!

2. Dec 3, 2005

### Pseudo Statistic

Is the first square root supposed to cover both 2 and y^2 or is it the square root of 2, plus y^2 (outside the root)?

3. Dec 3, 2005

### elle

oops sorry! Yes the square root is suppose to cover all of 2+y²

4. Dec 3, 2005

### Pseudo Statistic

In that case I'm guessing it's the same for the second square root term..
So you have:
∫{ (2y²)√(2+y²) - (2y²)√(2y²) } dy
Splitting it up into two integrals...
∫(2y²)√(2+y²) dy - ∫(2y²)√(2y²) dy
On the right side, you can pull the y^2 out of the square root:
∫(2y²)√(2+y²) dy - ∫(2y²)y√(2) dy
And then:
∫(2y²)√(2+y²) dy - 2√2 ∫y^3 dy
= ∫(2y²)√(2+y²) dy - (y^4)/(√2) + C
Now, for the left integral it'll take a bit more work.... let's try a hyperbolic substitution:
sinh^2 x - cosh^2 x = 1
sinh^2 x = 1 + cosh^2 x
So, let's say y = √2 * cosh x
dy = √2 * sinh x dx; from there:
∫(2y²)√(2+y²) dy - (y^4)/(√2) + C
= ∫(2(2cosh²x))√(2)*sinh x * √2 sinh x dx - (y^4)/(√2) + C
= 8∫cosh² x sinh² x dx - (y^4)/(√2) + C
And you can proceed from there.. :\ Or you could have used a trig substitution...

5. Dec 3, 2005

### elle

ooo ok thanks!

I've also got another quick question. I've been asked to draw the region of integration for the following integral. I'm not sure if I've drawn it right can someone help? thank you!

http://tinypic.com/i53khk.jpg"

Last edited by a moderator: Apr 21, 2017