Integrate cos^2x .
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We'll have to use the formula:
(cos x)^2 = [1 + cos(x/2)]/2
We'll integrate both sides:
Int (cos x)^2 dx = Int [1 + cos(x/2)]dx/2
We'll use the additive property of integral:
Int [1 + cos(x/2)]dx/2 = Int dx/2 + Int cos(x/2)dx/2
Int dx/2 = (1/2)/Int xdx
Int dx/2 = (x^2)/4 + C (1)
Int cos(x/2)dx/2 = (1/2)*Int cos(x/2)dx
(1/2)*Int cos(x/2)dx = (1/2)* sin(x/2)/(1/2) + C
(1/2)*Int cos(x/2)dx = sin(x/2) + C (2)
Int (cos x)^2 dx = (1) + (2)
Int (cos x)^2 dx = (x^2)/4 + sin(x/2) + C
To find Int cox^2x dx.
We know that cos2x = (cosx)^2 - (sin^x)^2
cos2x = (cosx)^2 -(1-coxx)^2
cos2x = 2(cosx)^2-1.
Add 1 and rewrite by changeing the sides:
Therefore (cosx)^2 = (1+cos2x)/2.
We replace (cosx)^2 by (1+cos2x)/2 in the given integral.
Therfore I = Int (cosx)^2 dx = Int (1+cos2x)dx
I = (1/2) Int (1+cos2x)d
I = (1/2)dx +(1/2) cos2xdx
I = (1/2)x + (1/2) (sin2x)/2
I = x/2 +(1/4) sin2x +constant.
Therefore Int (cosx)^2 dx = (x/2) +(1/4) sin2x + C
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