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

Answer

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

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