Question

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

Answer 1

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

Answer 2

Yes.

Answer 3

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

Answer 4

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

Answer 5

yep :)
now you can u-sub

Answer 6

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

Answer 7

or that^

Answer 8

Makes perfect sense.

Answer 9

Thank you both!

Answer 10

welcome :)

Answer 11

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

Answer 12

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