Asked by INTEGRAL

Please help..


ʃ (4sin²x cos²×/sin 2x cos 2x)dx

That's integration of (4sin^2x cos^2x over sin 2x cos 2x) dx

i've got it from the back and it has an answer from the back page of the book but i want to know how to solve it..i really need it for practicing this subject...please please help..Thanks a lot..
the answer is: -1/2 ln|secx + tanx| + c
Answered by Mgraph
4sin^2(x)cos^2(x)=(sin2x)^2
(sin2x/cos2x)dx=(-1/2)d(cos2x)/cos2x
Answered by Steve
Hmm. There's something wrong here. Mgraph's solution is correct, but is not what you were expecting.

ln|secx + tanx| is the integral of secx, not sinx/cosx

Is there a typo somewhere?
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