MCQ
$\int_{}^{} {\frac{{dx}}{{\tan x + \cot x}}} = $
  • A
    $\frac{{\cos 2x}}{4} + c$
  • B
    $\frac{{\sin 2x}}{4} + c$
  • C
    $ - \frac{{\sin 2x}}{4} + c$
  • $ - \frac{{\cos 2x}}{4} + c$

Answer

Correct option: D.
$ - \frac{{\cos 2x}}{4} + c$
d
(d)$\int_{}^{} {\frac{{dx}}{{\tan x + \cot x}}} = \int_{}^{} {\frac{{dx}}{{\frac{{\sin x}}{{\cos x}} + \frac{{\cos x}}{{\sin x}}}}} $
$ = \frac{1}{2}\int_{}^{} {2\sin x\cos x\,dx} = \frac{{ - \cos 2x}}{4} + c$.

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