MCQ
$\int_{}^{} {{e^x}{{\tan }^2}({e^x})dx = } $
  • A
    $\tan ({e^x}) - x + c$
  • B
    ${e^x}(\tan {e^x} - 1) + c$
  • C
    $\sec ({e^x}) + c$
  • $\tan ({e^x}) - {e^x} + c$

Answer

Correct option: D.
$\tan ({e^x}) - {e^x} + c$
d
(d) Put ${e^x} = t \Rightarrow {e^x}dx = dt,$ then
$\int_{}^{} {{e^x}{{\tan }^2}({e^x})\,dx} = \int_{}^{} {{{\tan }^2}t\,dt} = \int_{}^{} {({{\sec }^2}t - 1)\,dt} $
$ = \tan t - t + c = \tan ({e^x}) - {e^x} + c.$

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