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Math Help - Definite integration by reduction methods.

  1. #1
    Super Member Showcase_22's Avatar
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    Definite integration by reduction methods.



    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.....
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  2. #2
    Super Member Showcase_22's Avatar
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    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!

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  3. #3
    Super Member PaulRS's Avatar
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    2. Note that: <br />
I_n  + I_{n + 1}  = \int_0^1 {\tfrac{{\sinh ^{2n} \left( x \right)}}<br />
{{\cosh \left( x \right)}} \cdot \left[ {1 + \sinh ^2 \left( x \right)} \right]dx} <br />

    Recall the identity: <br />
1 + \sinh ^2 \left( x \right) = \cosh ^2 \left( x \right)<br />

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

    1. <br />
I_n  = \left( {\tfrac{{3n}}<br />
{{3n + 2}}} \right) \cdot I_{n - 1}  = \left( {\tfrac{{3n}}<br />
{{3n + 2}}} \right) \cdot \left( {\tfrac{{3\left( {n - 1} \right)}}<br />
{{3\left( {n - 1} \right) + 2}}} \right) \cdot I_{n - 2} <br />

    And so on: <br />
I_n  = \prod\limits_{k = 1}^n {\left( {\tfrac{{3k}}<br />
{{3k + 2}}} \right)}  \cdot I_0 <br />
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  4. #4
    Super Member Showcase_22's Avatar
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    Boy is my face red! I was seriously overcomplicating both questions!!
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  5. #5
    Super Member Showcase_22's Avatar
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    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.
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  6. #6
    Moo
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    Quote Originally Posted by Showcase_22 View Post
    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
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  7. #7
    Super Member Showcase_22's Avatar
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    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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