(sinx + cosx)^2 ?
There are limits given, pi/2 and 0, and the answer is 2.57.
Pls dun tell me to do my hw myself cos this is for enrichment.
Pls give detailed steps.
Thank you.
2006-09-05 02:22:10 · 6 answers · asked by Forest_aude 3 in Science & Mathematics ➔ Mathematics
Square it out to get sin^2 x +2sin x cos x +cos^2 x,
which is the same as 1+sin(2x). The anti-derivative of this is
x-(1/2)cos(2x). Evaluate this from x=0 to x=pi/2 to get an answer of
(pi +2)/2.
2006-09-05 03:37:42 · answer #1 · answered by mathematician 7 · 1⤊ 0⤋
first expand it
1+2sinxcosx = 1+ sin2x
integrate 1 from 0 to pi/2 it is pi/2
integrate sin2x from 0 to pi/2 it is -cos2x/2 from 0 to pi/2
it is (cos 0 - cos PI)/2 = 1
so the sum = 1+pi/2 put the value of Pi and you get the result
2006-09-05 10:40:09 · answer #2 · answered by Mein Hoon Na 7 · 0⤊ 0⤋
expand the (cos x + sin x)^2 expression.
then you get cos^2 x + sin^2 x + 2sin x.cos x = 1 +sin 2x
because cos^2 x + sin^2 x =1 and 2sin x.cos x = sin 2x
so integrating this we get [x - (1/2)cos 2x] from 0 to pi/2
= [ pi/2 - (1/2).(-1)] - [ 0 - (1/2)(1)]
= pi/2 + 1
So I reckon your answer is approximately right but I like to leave things in their precise form.
Hope this helps!
2006-09-05 12:27:44 · answer #3 · answered by yasiru89 6 · 0⤊ 0⤋
use wallis formula because theres a given limits 0 to pi/2
2006-09-05 10:03:23 · answer #4 · answered by jhen_hidaka 1 · 0⤊ 1⤋
Expand the equation first.
(sinx)^2 + 2sinxcosx + (cosx)^2
Using cos2x identities,
cos2x = 1 - 2(sinx)^2 = 2(cosx)^2 -1
Thus becomes,
(1-cos2x)/2 + sin2x + (cos2x+1)/2
I'm sure you can integrate from here. :)
2006-09-05 09:28:04 · answer #5 · answered by Anonymous · 0⤊ 1⤋
1st step buy a good calculater
2006-09-05 09:26:46 · answer #6 · answered by Jellybean 4 · 0⤊ 0⤋