Reduction formulae question?

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Think of the integral as 1*(1 - x^2)^n and use integration by parts with u = (1 - x^2)^n and dv = 1.

This leads to the indefinite integral
x(1 - x^2)^n - INT -2nx^2*(1 - x^2)^(n-1) dx

Evaluating between 0 and 1 shows that the first term is zero leaving
I(n) = 2n*INT [0, 1] (x^2)*(1 - x^2)^(n - 1) dx

Now change the x^2 into 1 - (1 - x^2) and separate it into two integrals. You will find that one of them can be expressed as a multiple of I(n).

Can you finish from there?...Show more

I(n) = ∫[0,1] ( 1 -x² )ⁿ dx ....... (1)
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Put x = sin u so that dx = cos u du.

Also, x in [0,1] means u in [0,π/2].
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From (1), then,

I(n) = ∫[0,π/2] ( cos²ⁿ u )· cos u du = *[0,π/2] cos²ⁿ⁺¹ u du ... ... (2)

Taking U = cos²ⁿ u and dV = cos u du,

and integrating by parts,

I(n) = { [ (cos²ⁿ u)·sin u ] on [0,π/2] } - ∫[0,π/2] ( 2n. cos²ⁿֿ¹ u.(- sin u ))· sin u du

. . . = [ 0 - 0 ] + 2n · ∫ cos²ⁿֿ¹ u. sin² u du

. . . = 2n · ∫ cos²ⁿֿ¹ u. ( 1 - cos² u ) du

. . . .= 2n · ∫ ( cos²ⁿֿ¹ u - cos²ⁿ⁺¹ u ) du

. . . .= 2n·I(n-1) - 2n·I(n) ....................... from (2)
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∴ I(n) = 2n·I(n-1) - 2n·I(n) ∴ (2n+1)·I(n) = 2n·I(n-1)

∴ I(n) = [ 2n / (2n+1) ]· I(n-1) ........................................ Q.E.D.
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Happy To Help !
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