calculus
posted by Dora .
evaluate the following integrals by any means possible:
b. integral from 1 to 1 (w/ (w^2+1)) (dw)
c. integral from 3 to 0 sqrt(9x^2)dx

b) zero because the function is odd.
c) substitute x = 3 sin(t)
Integral from pi/2 to 0 of
9 cos^2(t) dt
If you were to replace cos^2 by sin^2 in this integral, the answer would be the same because
cos(x) = sin(pi/2x) >
cos^2(x) = sin^(pi/2  x)
So, the values attained by cos^2 are also attained by sin^2 in the inteval from pi/2 to 0. If we call the integral I, then:
2 I = 9 Integral of
[cos^2(t) + sin^2(t)] dt from pi/2 to 0 =
9* pi/2 >
I = 9/4 pi