MCQ
$\int_{}^{} {\frac{{\cos 2x + 2{{\sin }^2}x}}{{{{\cos }^2}x}}dx = } $
  • A
    $2\sec x + c$
  • B
    $2\tan x + c$
  • $\tan x + c$
  • D
    None of these

Answer

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

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