# What is the integral of the function y=cos^2x-sin^2x?

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We have to find the integral of y = (cos x)^2 - (sin x)^2.

We know that cos 2x = (cos x)^2 - (sin x)^2

=> y = cos 2x

Int[ cos 2x dx]

let 2x = y, dy/2 = dx

=> Int[(1/2)cos y dy]

=> (1/2) sin y + C

substitute y = 2x

=> (1/2)*sin 2x + C

**The required integral is (1/2)*sin 2x + C**

To find the integral of cos^2x-sin^2x.

We know that cos^2x-sin^2x = cos2x.

=> Int (cos^2x-sin^2x) dx = Int cos2x dx

=> **Int (cos^2x-sin^2x) dx = (1/2)sin2x+C**.

We'll have to use the double angle identities:

(cos x)^2 = [1 + cos(x/2)]/2 (1)

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

We'll subtract (2) from (1):

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

We'll eliminate like terms:

(cos x)^2 - (sin x)^2 = cos(x/2)

We'll integrate both sides:

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

Int cos(x/2) dx = sin(x/2)/(1/2) + C

Int cos(x/2) dx = 2sin(x/2) + C

**The requested integral of the difference (cos x)^2 - (sin x)^2 is: Int [(cos x)^2 - (sin x)^2] dx = 2sin(x/2) + C.**