# calculus

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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(9-x^2)dx

• calculus -

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/2-x) ------->

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

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