MCQ
$\int_{}^{} {\frac{{{{\sin }^2}x - {{\cos }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}}dx = } $
  • $\tan x + \cot x + c$
  • B
    $\tan x + {\rm{cosec}}\,x + c$
  • C
    $ - \tan x + \cot x + c$
  • D
    $\tan x + \sec x + c$

Answer

Correct option: A.
$\tan x + \cot x + c$
a
(a)$\int_{}^{} {\frac{{{{\sin }^2}x - {{\cos }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}}} \,dx = \int_{}^{} {{{\sec }^2}x\,dx - \int_{}^{} {{\rm{cose}}{{\rm{c}}^{\rm{2}}}\,x\,dx} } $
$ = \tan x + \cot x + c.$

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