Question

the integral of 16sin^2xcos^2xdx
it needs to be solved using trig substitution.

Answer 1

is this substitution \(2 \sin x \cos x = \sin 2x\) allowed?

Answer 2

\[\int\limits 16\sin^2x \cos^2x \space dx\]\[\sin^2x={1-\cos(2x) \over 2}\]\[\cos^2x={1+\cos(2x) \over 2}\]\[\int\limits 16{1-\cos(2x) \over 2} {1+\cos(2x) \over 2} \space dx\]\[\int\limits\limits 4(1-\cos(2x))( 1+\cos(2x)) \space dx\]\[\int\limits\limits 4(1-cos^2(2x)) \space dx\]\[u=2x\]\[{du \over 2}=dx\]\[\sin^2u=1-\cos^2u\]\[\int\limits 2\sin^2u \space du\]\[2({1 \over 2}u-{1 \over 4}\sin2u)+C\]\[u-{1 \over 2}\sin2u+C\]\[2x-{1 \over 2}\sin4x+C\]

Answer 3

\[\int\limits \sin^2u \space du={1 \over 2}u-{1 \over 4}\sin2u+C\]