How to integrate [sin(2*pi*x)]^2 ?

Answer 1

Integral (sin^2[2pi(x)]) dx

To solve this integral, you use the half angle identity:

sin^2(y) = (1/2)(1 - cos2y)

In our case, y = 2pi(x), so our integral becomes

Integral ( [1/2] [1 - cos(2[2pi(x)]) ) dx

Pull the constant (1/2) out of the integral, to get

(1/2) * Integral (1 - cos (2[2pi(x)]) ) dx

Reducing the stuff inside of the cosine,

(1/2) * Integral (1 - cos [4pi(x)] ) dx

And now, we have a linear function inside of the cosine, making it easily integrable. All we have to do is divide by 4pi to offset the chain rule. Our answer is

(1/2) * [x - (1/(4pi)) sin(4pi(x))] + C

Simplifying,

(1/2)x - (1/[8pi]) sin[4pi(x)] + C

Answer 2

Note that cos 2A = cos^2 A - sin^2 A = 1 - 2 sin^2 A.
So sin^2 (2πx) = (1 - cos (4πx))/2.

Hence ∫ sin^2 (2πx) dx = ∫ (1 - cos (4πx))/2 dx
= x/2 - sin (4πx) / (2.(4π)) + c
= x/2 - sin (4πx) / 8π + c.

Answer 3

x/2 - sin(4*pi*x)/(8*pi) +c if you need integral formilas go to: http://www.sosmath.com/tables/integral/integ19/integ19.html

Answer 4

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