Math Is Fun Forum

Al-Allo

Member

Registered: 2012-08-23

Posts: 324

Integrate ∫(1+4y^2)^(1/2) dy from 0 to 1

I have the following integral to solve :

http://www.wolframalpha.com/input/?i=in … rom+0+to+1

I was able to integrate and I got : (y/4)+((e^(2y)-e^(-2y))/16)

I don't what to do after that. I know I can replace ((e2y-e-2y)/2) with sinh(2y)... but after that I'm lost. Help please ? Thanks

I'm using the following substitution : y=sinhθ/2

Last edited by Al-Allo (2015-04-26 18:12:57)

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Integrate ∫(1+4y^2)^(1/2) dy from 0 to 1

Hi;

Is that e^(2y) in (y/4)+((e2y-e-2y)/16)?

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Al-Allo

Member

Registered: 2012-08-23

Posts: 324

Re: Integrate ∫(1+4y^2)^(1/2) dy from 0 to 1

Yes, sorry it was an error. I edited. Anyway, Im going to go sleep.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Integrate ∫(1+4y^2)^(1/2) dy from 0 to 1

hi Al-Allo

I got that too.

You can write

You know sinh(y) = 2 and you can work out cosh(y) using 1 + sinh^2 = cosh^2

Hope that helps,

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob