Wolfram said this:
http://www.wolframalpha.com/input/?i=integral+cos+2x+ln%28sin+2x%29+dx
look at the last version of the "alternate forms"
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!
1 answer
4 answers