# When integrate sin x gives -cosx. When integrate sin^2x gives cos^2x.

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### 2 Answers

1)

The idefinite integral of sinx dx = -cosx .

2)

Int sin^2 dx = Int (1-cos2x)/2 dx.

Int sin^2 dx = Int (1/2)dx - (1/2)Intcos2x

Int sin^2x dx = x/2 -(1/2) (sin2x)/2

Int sin^2dx = x/2 -(1/4)sin2x.

Threfore Int sin^2xdx could be equal to cos^2 x if x/2 -(1/4)sin2x = cos^2x.

We'll have to use the formula:

(sin x)^2 = [1 - cos(x/2)]/2

We'll integrate both sides:

Int (sin 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 (sin x)^2 dx = (1) + (2)

**Int (sin x)^2 dx = (x^2)/4 + sin(x/2) + C**

**So, when integrate (sinx)^2, the result is not (cosx)^2.**