(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)].
1 answer:
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