Best Answer:
Let J(n) = ∫(0 to π/2) sin^(2n)(x) dx.

J(1) = ∫(0 to π/2) sin^2(x) dx

= ∫(0 to π/2) (1/2)(1 - cos(2x)) dx

= (1/2)(x - sin(2x)/2) {for x = 0 to π/2}

= (1/2)(π/2).

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For n > 1, note that ∫(0 to π/2) sin^(2n)(x) dx = ∫(0 to π/2) sin^(2n-1)(x) * sin x dx.

Use integration by parts with u = sin^(2n-1)(x), dv = sin x dx

and du = (2n-1) sin^(2n-2)(x) cos x dx, v = -cos x.

So, we obtain ∫(0 to π/2) sin^(2n)(x) dx

= -sin^(2n-1)(x) cos x {for x = 0 to π/2} - ∫(0 to π/2) -(2n-1) sin^(2n-2)(x) cos^2(x) dx

= 0 + (2n-1) ∫(0 to π/2) sin^(2n-2)(x) (1 - sin^2(x)) dx

= (2n-1) ∫(0 to π/2) [sin^(2n-2)(x) - sin^(2n)(x)] dx

In summary, we have

∫(0 to π/2) sin^(2n)(x) dx = (2n-1) ∫(0 to π/2) [sin^(2(n-1))(x) - sin^(2n)(x)] dx

==> J(n) = (2n - 1) [J(n-1) - J(n)].

Solving for J(n) yields

J(n) = [(2n - 1)/(2n)] J(n-1).

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Now, we just repeatedly use this relation, until we get to J(1).

J(n) = [(2n - 1)/(2n)] J(n-1)

.......= [(2n - 1)/(2n)] * [(2n - 3)/(2n - 2)] J(n-2)

.......= [(2n - 1)(2n - 3)/((2n)(2n - 2))] * [(2n - 5)/(2n - 4)] J(n-3)

...

.......= [((2n - 1)(2n - 3)...3) / ((2n)(2n - 2)...4)] J(1)

.......= [((2n - 1)(2n - 3)...3) / ((2n)(2n - 2)...4)] * (1/2)(π/2)

.......= [((2n - 1)(2n - 3)...3 * 1) / ((2n)(2n - 2)...4 * 2)] (π/2), as required.

I hope this helps!

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