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integrate cos2x/(cos^2x*sin^2x)
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\[\int{\cos(2x)\over\cos^2x\sin^2x}dx\]?
Yes.
hint\[\cos x\sin x=\frac12\sin(2x)\]
Therefore cos^2x*sin^2x=1/4sin^2(2x)?
yep :) now you can u-sub
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cos2x=cos^2x -sin^2x thus cos2x/(cos^2x*sin^2x)=(cos^2x -sin^2x)/(cos^2x*sin^2x) =cosec^2x -sec^2x hnce required integral equals -cotx -secx +c
or that^
Makes perfect sense.
Thank you both!
welcome :)
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welcome @moderator i think easy answers and complete answers are required more.....wasting time ..
@moderator i don't care if i recieve medals or not...
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