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.....
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.....
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
$