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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)?

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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: 10,619

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

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