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# reduction formula questions watch

1. i have been trying to do this questions but i cannot solve them. please help me! the questions are from the review exercise of the Heinemann Pure Mathematics 5 textbook.
Question 50, part c and d. (I can do a and b ok), and
Question 61, part a (i could probably do part b and c because the formula is given, but because i haven't been able to derive it i haven't gotten to solve b and c).
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2. 50. I(n) = [(n-1)/n] I(n-2)
(c)
I(2n) = [(2n-1)/(2n)] I(2n-2)
= [(2n-1)/(2n)] [(2n-3)/(2n-2)] I(2n-4)
= [(2n-1)/(2n)] [(2n-3)/(2n-2)] [(2n-5)/(2n-4)] ... I(0)

Now, I(0) = pi/2. So:
I(2n) = [(2n-1)/(2n)] [(2n-3)/(2n-2)] [(2n-5)/(2n-4)] ... pi/2
= [1/2^(n)] . [(2n-1)(2n-3)(2n-5)...]/[n(n-1)(n-2)...] . pi/2
= [1/2^(n+1)] . [(2n-1)(2n-3)(2n-5)...]/(n!) . pi

You can re-write (2n-1)(2n-3)... as (2n!)/[(2n-2)(2n-4)...] = (2n!)/[2^(n) (n!)]. So:
I(2n) = [1/2^(n+1)] . (2n!)/[2^(n) (n!)²] . pi
= [(2n!)pi]/[2^(2n+1) (n!)²], as required.

(d) Do the same for I(2n-1), then multiply it by the expression for I(2n).

61. I(n) = 01 1/(x²+a²)^n dx
(a) Integrate by parts with:
u = (x²+a²)^(-n), dv/dx=1
du/dx = -2nx.(x²+a²)^(-n-1), v=x

Then:
I(n) = [x/(x²+a²)^n]{0, 1} + 2n 01 x²/(x²+a²)^(n+1) dx
= 1/(1+a²)^n + 2n 01 (x²+a²-a²)/(x²+a²)^(n+1) dx
= 1/(1+a²)^n + 2n 01 (x²+a²)/(x²+a²)^(n+1) dx - 2na² 01 1/(x²+a²)^(n+1)
= 1/(1+a²)^n + 2n 01 1/(x²+a²)^n dx - 2na².I(n+1)
= 1/(1+a²)^n + 2n.I(n) - 2na².I(n+1)

So:
2na² I(n+1) = 1/(1+a²)^n + 2n I(n) - I(n)
= 1/(1+a²)^n + (2n-1) I(n), as required.
3. i listend to some music and then came back to the problem and i solved it. here is the solution in two parts. thank you for the answer to 50.
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