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    • Thread Starter

    Hi there,
    would anyone be able to give me a hand with question as I am unsure what to do. Thank you
    • Thread Starter


    its question 69 thanks x

    With these trig types it is usually best to start off by trying

     \displaystye I_n =\int_{0}^{ \pi}e^x\sin^n x dx = \int_{0}^{ \pi} e^x\sin^{n-2} x\sin^2 x dx .

    How do this help you say, well..

     I_n =\displaystyle \int_{0}^{ \pi} e^x\sin^{n-2} x(1-\cos^2 x) dx which you can then expand leaving you with 2 terms, one of them being  I_{n-2} .
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