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

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Answer 1 · 2016-02-05T17:32:35

We have that

$cos(4x)=cos^2(2x)-sin^2(2x)=>$
$cos(4x)=1-sin^2(2x)-sin^2(2x)=>$
$2sin^2(2x)=1-cos(4x)=>$
$sin^2(2x)=1/2*(1-cos(4x))$

Hence we have that

$int sin^2(2x)dx=int [1/2*(1-cos(4x))]dx=x/2-sin(4x)/8+c$

Footnote

We used the following trig identities

1) $cos(2*a)=cos^2a-sin^2a$

2) $cos^2a=1-sin^2a$

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