Question

integral of sec^2xcosec^2x dx

Answer 1

sec²x.cosec²x .dx

=dx/cos²x.sin²x

we know,
sin²x +cos²x =1
put this ,

(sin²x +cos²x)dx/sin²x.cos²x

=sec²x.dx + cosec²x.dx

=tanx -cotx

=tanx -1/tanx

=tan²x -1/tanx

= -{2/2tanx/(1-tan²x)}

use formula ,
tan2x =2tanx/(1-tan²x)

=-2{1/tan2x}

=-2cot2x +C

where C is constant .