What is the integral of $int (cos2x)^2 dx$ ?

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Answer 1 · 2016-07-20T10:23:35

$= 1/8 sin 4x + x/2 + C$

use the ID: $cos 2A = 2 cos^2 A - 1$ or $cos^2 A = 1/2(cos 2A + 1)$

the integral becomes

$1/2 int cos4x + 1 \ dx$

$= 1/2 (1/4 sin 4x + x) + C$

$= 1/8 sin 4x + x/2 + C$

What is the integral of int (cos2x)^2 dxint (cos2x)^2 dx?