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    • Thread Starter

    Hi folks.

    I'm just getting started with the work here and I'm beginning to think my gap year's taken its toll, as I can't wrap my head around a couple of 'warm up' questions for our first supervision . I think I'll get back into my stride eventually but for now I'd like to avoid looking a dunce in my first supervision if possible.

    I don't necessarily want full solutions as I do want to be able to work these out myself, but I'd appreciate any hints or pointers.

    Using integration by parts, show that

    ∫sechn+2u.sinhu.sinhu du = (n+1)-1∫sechnu du
    (limits of integration between 0 and infinity)

    I can't really see where the (n+1)-1 would come from in this.


    An = (qn / n!)∫[x(π-x)]nsinxdx
    (limits of integration between 0 and pi)

    Show that An = (4n-2)qAn-1 - (qπ)2An-2

    Thanks again for any help. I don't have a whole lot to give but rep is available.

    \int sech^{n+2}u \, sinh u du = \frac{1}{n+1}sech^{n+1}u

    That's using x = cosh u. Use that result to get the first question.
    • Thread Starter

    Ah I see. I should've spotted that the 1/(n+1) might come from integrating something with a (n+1) power, so it was just a matter of finding a substitution that allowed me to integrate the multiple sech's.

    Thanks AN, that worked a treat, rep on way as soon as I'm able to give it.
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Updated: October 8, 2006

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