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FP3 Intergration

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.
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 (n1)(n-1) can be put outside, and that xn2(25x2)=25xn2xnx^{n-2} (25 - x^2) = 25x^{n-2} - x^n

This means the integral can be rewritten as

(n1)0525xn2xn25x2.dx\displaystyle (n-1) \int_0^5 \dfrac{25x^{n-2}-x^n}{\sqrt{25-x^2}}.dx

=25(n1)05xn225x2.dx(n1)05xn25x2.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 In2I_{n-2} and the second is InI_n
(edited 6 years ago)
Reply 2
Original post by RDKGames
Note that (n1)(n-1) can be put outside, and that xn2(25x2)=25xn2xnx^{n-2} (25 - x^2) = 25x^{n-2} - x^n

This means the integral can be rewritten as

(n1)0525xn2xn25x2.dx\displaystyle (n-1) \int_0^5 \dfrac{25x^{n-2}-x^n}{\sqrt{25-x^2}}.dx

=25(n1)05xn225x2.dx(n1)05xn25x2.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 In2I_{n-2} and the second is InI_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 :biggrin:

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