# Definite integration by reduction methods.

• Aug 30th 2008, 03:30 AM
Showcase_22
Definite integration by reduction methods.
http://i116.photobucket.com/albums/o...thsproblem.jpg

There's my problem. I managed to solve the first part but i'm utterly stumped on the last bit. I think it has something to do with making another expression for In-1 and substituting it in. I haven't been able to it though.....
• Aug 30th 2008, 04:18 AM
Showcase_22
After posting this thread I found another question that I couldn't do. I've done as much of it as I can and I think i'm missing a trick of some sort. Solving this one and the question before would be great!

http://i116.photobucket.com/albums/o...hsproblem1.jpg
• Aug 30th 2008, 05:02 AM
PaulRS
2. Note that: $\displaystyle I_n + I_{n + 1} = \int_0^1 {\tfrac{{\sinh ^{2n} \left( x \right)}} {{\cosh \left( x \right)}} \cdot \left[ {1 + \sinh ^2 \left( x \right)} \right]dx}$

Recall the identity: $\displaystyle 1 + \sinh ^2 \left( x \right) = \cosh ^2 \left( x \right)$

So that: $\displaystyle I_n + I_{n + 1} = \int_0^1 {\sinh ^{2n} \left( x \right)\cosh \left( x \right)dx}$ now let $\displaystyle u = \sinh \left( x \right)$ and we are done

1. $\displaystyle I_n = \left( {\tfrac{{3n}} {{3n + 2}}} \right) \cdot I_{n - 1} = \left( {\tfrac{{3n}} {{3n + 2}}} \right) \cdot \left( {\tfrac{{3\left( {n - 1} \right)}} {{3\left( {n - 1} \right) + 2}}} \right) \cdot I_{n - 2}$

And so on: $\displaystyle I_n = \prod\limits_{k = 1}^n {\left( {\tfrac{{3k}} {{3k + 2}}} \right)} \cdot I_0$
• Aug 30th 2008, 05:20 AM
Showcase_22
Boy is my face red! I was seriously overcomplicating both questions!!
• Aug 30th 2008, 05:34 AM
Showcase_22
How do you show that Io is -(pi/2)+2arctan e? I'm trying to do it and my method of substitution isn't working.(Crying)
• Aug 30th 2008, 05:45 AM
Moo
Quote:

Originally Posted by Showcase_22
How do you show that Io is -(pi/2)+2arctan e? I'm trying to do it and my method of substitution isn't working.(Crying)

$\displaystyle I_0=\int_0^1 \frac{1}{\cosh(x)} ~dx$

$\displaystyle \cosh(x)=\frac{e^x+e^{-x}}{2}$

$\displaystyle \implies I_0=\int_0^1 \frac{2}{e^x+e^{-x}} ~dx=2 \int_0^1 \frac{e^x}{e^{2x}+1} ~dx$

Substitute $\displaystyle t=e^x$ :)
• Aug 31st 2008, 01:33 AM
Showcase_22
ohhhh, I was trying to keep it in terms of cosh which led to an amazing number of substitutions until I ended up back where I started!(Headbang)