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    I don't understand after breaking up sqrt(25-x^2) and all that how it integrates into In= 25(n-1)In-2 - (n-1)In
    What is it i'm not seeing?
    Sorry i don't know how to types maths haha.
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    (Original post by I'msoPi)
    I don't understand after breaking up sqrt(25-x^2) and all that how it integrates into In= 25(n-1)In-2 - (n-1)In
    What is it i'm not seeing?
    Sorry i don't know how to types maths haha.
    Note that (n-1) can be put outside, and that x^{n-2} (25 - x^2) = 25x^{n-2} - x^n

    This means the integral can be rewritten as

    \displaystyle (n-1) \int_0^5 \dfrac{25x^{n-2}-x^n}{\sqrt{25-x^2}}.dx

    \displaystyle = 25(n-1) \int_0^5 \dfrac{x^{n-2}}{\sqrt{25-x^2}} .dx - (n-1) \int_0^5 \dfrac{x^n}{\sqrt{25-x^2}}.dx

    The first integral is I_{n-2} and the second is I_n
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    (Original post by RDKGames)
    Note that (n-1) can be put outside, and that x^{n-2} (25 - x^2) = 25x^{n-2} - x^n

    This means the integral can be rewritten as

    \displaystyle (n-1) \int_0^5 \dfrac{25x^{n-2}-x^n}{\sqrt{25-x^2}}.dx

    \displaystyle = 25(n-1) \int_0^5 \dfrac{x^{n-2}}{\sqrt{25-x^2}} .dx - (n-1) \int_0^5 \dfrac{x^n}{\sqrt{25-x^2}}.dx

    The first integral is I_{n-2} and the second is I_n
    Yeah I put that (n-1) outside too, just makes life easier haha. Thank you so much for breaking it down for me, I was stumped for quite some time. It's clear as day now. Thanks again
 
 
 
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