Question

(a) Use the reduction formula to show that integral from 0 to pi/2 of sin(x)^ndx is (n-1)/n * integral from 0 to pi/2 of sin(x)^(n-2)dx where n>=2 is an integer.

(b)Use part (a) to evaluate integral from 0 to pi/2 of sin(x)^3dx and integral from 0 to pi/2 of sin(x)^5dx.

(c) Use part (a) to show that, for odd powers of sine, integral from 0 to pi/2 sin(x)^(2n+1)dx is (2*4*6*...*2n)/[3*5*7...*(2n+1)].

Answer 1

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