MCQ
$\int_{}^{} {\frac{{\cot x\tan x}}{{{{\sec }^2}x - 1}}} \;dx = $
  • A
    $\cot x - x + c$
  • B
    $ - \cot x + x + c$
  • C
    $\cot x + x + c$
  • $ - \cot x - x + c$

Answer

Correct option: D.
$ - \cot x - x + c$
d
(d) $\int_{}^{} {\frac{{\cot x\tan x}}{{{{\sec }^2}x - 1}}\,dx = \int_{}^{} {\frac{1}{{{{\tan }^2}x}}\,dx = \int_{}^{} {{{\cot }^2}x\,dx} } } $$ = \int_{}^{} {(co{\rm{se}}{{\rm{c}}^2}x - 1)\,dx} = - \cot x - x + c.$

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