HSC Stuff > HSC Mathematics Extension 2

Recurrence Relations

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annaconda:
Hello all,

Since the 4U exam is in 10 days (lol rip), I was still wondering, how can I easily tell how to split the integral to perform IBP without it getting to overly complicated?

An example would be HSC 2013, Q13 a) part i

Thanks :)

RuiAce:
It's hard to say. For that particular one, the way to go about the IBP is essentially the only way, which is to treat your \(u\) being literally everything that's there and then integrating \(1\) back into \(x\).
\[ I_n = \left[x(1-x^2)^{n/2} \right]_0^1 - \int_0^1x \cdot -2x \cdot \frac{n}{2} (1-x^2)^{n/2-1}\,dx \]
The one thing to note for recurrence relations in integrals is that there are two important things. The first is obviously the need for IBP, but the second is the classic add/subtract the same thing, or multiply/divide the same thing, in order to simplify the expression. Usually if IBP is insufficient, and only leads to something even messier, that add/take-away trick becomes very important.

(Note that the square bracket term there evaluates to 0.)
\begin{align*}I_n &= -n \int_0^1 -x^2 (1-x^2)^{n/2-1}\,dx\\ &= -n\int_0^1 \left( (1-x^2) + 1 \right) (1-x^2)^{n/2-1}\\ &= -n \int_0^1 (1-x^2)^{n/2-1}(1-x^2) - (1-x^2)^{n/2-1} \, dx \tag{carefully expanding}\\ &= -n \int_0^1 (1-x^2)^{n/2} + n\int_0^1 (1-x^2)^{n/2 - 1}\,dx \end{align*}

annaconda:
yes omg thankyou !!!

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