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

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**