What is the integral of $int [sin(2pix)]^2$ ?

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Answer 1 · 2017-02-03T23:22:27

$int sin^2(2pix)dx = x/2 - 1/(8pi)sin(4pix) + C$

Use the trigonometric identity:

$sin^2theta = (1-cos2theta)/2$

We have:

$int sin^2(2pix)dx = int (1-cos4pix)/2dx$

using linearity:

$int sin^2(2pix)dx = 1/2 int dx -1/2 int cos4pixdx = x/2 - 1/(8pi) int cos(4pix)d(4pix) = x/2 - 1/(8pi)sin(4pix) + C$

What is the integral of int [sin(2*pi*x)]^2int [sin(2*pi*x)]^2?