How do you integrate #int sin^2(2x)dx#?

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Answer 1 · 2017-03-15T17:49:05

#1/8(4-sin(4x))+C#

Use the cosine double-angle identity to rewrite the function:

#cos(2alpha)=1-2sin^2(alpha)" "=>" "sin^2(alpha)=(1-cos(2alpha))/2#

Then:

#sin^2(2x)=(1-cos(4x))/2#

So:

#intsin^2(2x)dx=1/2int(1-cos(4x))dx=1/2intdx-1/2intcos(4x)dx#

The second can be solved with the substitution #u=4x=>du=4dx#:

#intsin^2(2x)dx=1/2x-1/8int4cos(4x)dx=1/2x-1/8intcos(u)du#

The integral of cosine is sine:

#intsin^2(2x)dx=1/2-1/8sin(u)=1/8(4-sin(u))=1/8(4-sin(4x))+C#