If you cannot answer that question, then I would be satisfied with just the antiderivative of (sin x)^2
2007-11-21 10:17:30 · 5 answers · asked by B.B. 3 in Science & Mathematics ➔ Mathematics
(1/2)( x - sin(x)cos(x) ) [2,0]
(1/2)( 2 - sin(2)cos(2) ) - (1/2)( 0 - sin(0)cos(0) )
1 - sin(2)cos(2)/2
2007-11-21 10:29:34 · answer #1 · answered by Anonymous · 0⤊ 0⤋
I agreed with Don fernando
Just substitute (sinx)^2=(1-cos2x)/2
u can get this from double angle trigo of cos2x
then simply integrate (1-cos2x)/2
2007-11-21 10:34:41 · answer #2 · answered by asker 1 · 0⤊ 0⤋
Its not too bad, its a trig identity issue:
Substitute (1-cos2x)/2 for (sinx)^2
That might help.
2007-11-21 10:28:13 · answer #3 · answered by Don Fernando 3 · 1⤊ 0⤋
S((sin x)^2,x)=
x/2-[sinXcosX]/2
from zero to 2 would depend if you are in degrees or rads. I hope you have it from here.
2007-11-21 10:30:44 · answer #4 · answered by info2know 3 · 0⤊ 0⤋
(sinx)^2 = (1/2)(1 - cos2x)
integ{(1/2)(1-cos2x)}=
integ(1/2) - (1/2)integ(cos2x)
=(1/2)x - (1/4)sin2x
=1-(1/4)sin4
2007-11-21 10:31:03 · answer #5 · answered by tinhnghichtlmt 3 · 0⤊ 0⤋