Posted by **christine** on Sunday, February 18, 2007 at 2:19pm.

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 dont 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.

## Answer This Question

## Related Questions

- TRIG! - Posted by hayden on Monday, February 23, 2009 at 4:05pm. sin^6 x + cos^6...
- Integral - That's the same as the integral of sin^2 x dx. Use integration by ...
- Integration by Parts - integral from 0 to 2pi of isin(t)e^(it)dt. I know my ...
- trig integration - s- integral endpoints are 0 and pi/2 i need to find the ...
- tigonometry - expres the following as sums and differences of sines or cosines ...
- trig - Reduce the following to the sine or cosine of one angle: (i) sin145*cos75...
- algebra - Can someone please help me do this problem? That would be great! ...
- Mathematics - Trigonometric Identities - Let y represent theta Prove: 1 + 1/tan^...
- Precal - I do not understand how to do this problem ((sin^3 A + cos^3 A)/(sin A...
- Trig - Find sin(s+t) and (s-t) if cos(s)= 1/5 and sin(t) = 3/5 and s and t are ...

More Related Questions