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Mathematics 20 Online
OpenStudy (anonymous):

integrate cos2x/(cos^2x*sin^2x)

OpenStudy (turingtest):

\[\int{\cos(2x)\over\cos^2x\sin^2x}dx\]?

OpenStudy (anonymous):

Yes.

OpenStudy (turingtest):

hint\[\cos x\sin x=\frac12\sin(2x)\]

OpenStudy (anonymous):

Therefore cos^2x*sin^2x=1/4sin^2(2x)?

OpenStudy (turingtest):

yep :) now you can u-sub

OpenStudy (anonymous):

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

OpenStudy (turingtest):

or that^

OpenStudy (anonymous):

Makes perfect sense.

OpenStudy (anonymous):

Thank you both!

OpenStudy (turingtest):

welcome :)

OpenStudy (anonymous):

welcome @moderator i think easy answers and complete answers are required more.....wasting time ..

OpenStudy (anonymous):

@moderator i don't care if i recieve medals or not...

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