MCQ
$\int_{}^{} {\frac{{\tan (\log x)}}{x}\;dx = } $
  • A
    $\log \cos (\log x) + c$
  • B
    $\log \sin (\log x) + c$
  • $\log \sec (\log x) + c$
  • D
    $\log {\rm{cosec}}(\log x) + c$

Answer

Correct option: C.
$\log \sec (\log x) + c$
(c) Put $\log x = t \Rightarrow \frac{1}{x}dx = dt$, therefore
$\int_{}^{} {\frac{{\tan (\log x)}}{x}dx = \int_{}^{} {\tan t\;dt} } $
$ = \log \sec t + c = \log \;\sec (\log x) + c$.

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