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1. anyone pls. tell me how to intergrate this

given that

∫(1-x²)^n dx, the limits are 0 to 1

prove that

(2n+1)In = 2nI(n-1)

would be grateful.thanks

all the best to all.
2. I(n)
= (1 - x^2)^n dx
= (1 - x^2)(1 - x^2)^(n - 1) dx
= I(n - 1) - x^2(1 - x^2)^(n - 1) dx

--

x^2(1 - x^2)^(n - 1) dx
= x * x(1 - x^2)^(n - 1) dx
= [ x * (-1/(2n))(1 - x^2)^n ](from 0 to 1) - (-1/(2n))(1 - x^2)^n dx
= 0 + I(n)/(2n)
= I(n)/(2n)

--

I(n) = I(n - 1) - I(n)/(2n)
2n I(n) = 2n I(n - 1) - I(n)
(2n + 1)I(n) = 2n I(n - 1)

--

3. thanks you are a real life saver. gave me a new hope of light. TODAY is the monster
4. (Original post by ruzaika)
anyone pls. tell me how to intergrate this

given that

∫(1-x²)^n dx, the limits are 0 to 1

prove that

(2n+1)In = 2nI(n-1)

would be grateful.thanks

all the best to all.
very nice question - thanks for posting that
5. I thought of a slightly neater solution.

I(n)
= (1 - x^2)^n dx
= 1*(1 - x^2)^n dx
= [ x*(1 - x^2)^n ](from 0 to 1) + 2n x^2 (1 - x^2)^(n - 1) dx
= 0 + 2n (1 - (1 - x^2)) (1 - x^2)^(n - 1) dx
= 2n [I(n - 1) - I(n)]

(2n + 1)I(n) = 2n I(n - 1)

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