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

Question

108983 views around the world

You can reuse this answer
Creative Commons License

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#