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Integral of cos^2x=(1/2)(cosxsinx+x)+CHere is why:Here is one method: use integration by parts and let u=cosx and dv=cosxdxdu=-sinx v=sinxInt(udv)=uv-Int(vdu) so uv=cosx(sinx) and vdu=sinx(-sinx)so we have:Int(cos^2(x)=(cosx)(sinx)+Int(sin^2x)(the (-) became + because of the -sinx, so we add Int(vdu))Now it looks not better because we have sin^2x instead of cos^2x,but sin^2x=1-cos^2x since sin^2x+cos^2x=1So we haveInt(cos^2x)=cosxsinx+Int(1-cos^2x)=cosxsinx+Int(1)-Int(cos^2x)So now add the -Int(cos^2x) on the RHS to the one on the LHS2Int(cos^2x)=cosxsinx+xso Int(cos^2x)=1/2[cosxsinsx+x] and now add the constant!final answerIntegral of cos^2x=(1/2)(cosx sinx + x)+C = x/2 + (1/4)sin 2x + C(because sin x cos x = (1/2)sin 2x)Another method is:Use the half-angle identity, (cos x)^2 = (1/2)(1 + cos 2x). So we have:Int[(cos x)^2 dx] = Int[(1/2)(1 + cos 2x)] dx = (1/2)[[Int(1 dx)] + [Int(cos 2x dx)]]= (1/2)[x + (1/2)sin 2x] + C= x/2 +(1/4)sin 2x + C
If x equals 10 and y equals 10, then 9x plus 8y equals 170.
Find the measure of this angles m1 equals 123 m8 equals?
The second would suggest that x equals the string "=3".
2
y=sinx y=cosxsinx=cosx=>sinx/cosx=1=>tanx=1=>x=45oie.. y=sin45=cos45y=1/(square root of 2)
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Without specifying the limits of integration, the integral will always include an arbitrary constant, and you'll never get a numerical value for it. So the statement is untrue on its face, hence unprovable. We'll have a look at the trig later.
You know that cosπ/4 and sinπ/4 both equal 1/root2, so multiplying them gets 1/2.
YES!!!! Sin(2x) = Sin(x+x') Sin(x+x') = SinxCosx' + CosxSinx' I have put a 'dash' on an 'x' only to show its position in the identity. Both x & x' carry the same value. Hence SinxCosx' + CosxSinx' = Sinx Cos x + Sinx'Cosx => 2SinxCosx
one equals
That 8 might equal D. Trashed. "Eight Equals D" is: 8=D Which is a penis smiley -.- ~ Skullcandy88
equals D repeater
(It's a penis) 8=====D 8 = balls D = head
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