Calculus

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Evaluate using integration by parts, substitution, or both if necessary.

the intergral of cos 2x ln(sin 2x) dx

My work:

w= sin2x
dw= 2cos2xdx
1/2 dw= cos2xdx

1/2 integrsl sign ln(w)dw

u= lnw
u'= 1/w
v= w
v'=1

1/2 [(lnw)(w)- integral sign (1/w)(w) dw]
1/2 (wlnw-w)

Final Answer:
1/2 sin2xln(sin2x)-1/2 sin2x

This answer seems correct to me, but when I typed this in, the answer is said to be incorrect!

Please check my work and see if I made any mistakes.
Thank you!

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