trig integration

i'm having trouble evaluating the integral at pi/2 and 0.

i know:

s (at pi/2 and 0) sin^2 (2x)dx=
s 1/2[1-cos(2x)]dx=
s 1/2(x-sin(4x))dx=
(x/2)- 1/8[sin (4x)]

i don't understand how you get pi/4

You made a few mistakes, check again. But you don't need to do any calculations to find out that the integral is pi/4. You just note that cos^(2x) integrated over the same interval will yield the same value, because it attains the same values in that interval as sin^(2x) does.

If you add up the two integrals then, because sin^2 + cos^= 1, you get pi/2. SO each integral is pi/4.

asked by christine

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