Definite integration by reduction methods.

**Showcase_22** · August 30th 2008, 03:30 AM

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.....

**Showcase_22** · August 30th 2008, 04:18 AM

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!

**PaulRS** · August 30th 2008, 05:02 AM

2. Note that: $ 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: $ 1 + \sinh ^2 \left( x \right) = \cosh ^2 \left( x \right) $

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

1. $ 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: $ I_n = \prod\limits_{k = 1}^n {\left( {\tfrac{{3k}} {{3k + 2}}} \right)} \cdot I_0 $

**Showcase_22** · August 30th 2008, 05:20 AM

Boy is my face red! I was seriously overcomplicating both questions!!

**Showcase_22** · August 30th 2008, 05:34 AM

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.

**Moo** · August 30th 2008, 05:45 AM

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.

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

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

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

Substitute $t=e^x$

**Showcase_22** · August 31st 2008, 01:33 AM

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!

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