MCQ
$\int_{}^{} {\sec x\log (\sec x + \tan x)\;dx = } $
  • A
    ${[\log (\sec x + \tan x)]^2} + c$
  • $\frac{1}{2}{[\log (\sec x + \tan x)]^2} + c$
  • C
    ${\sec ^2}x + \tan x\sec x + c$
  • D
    None of these

Answer

Correct option: B.
$\frac{1}{2}{[\log (\sec x + \tan x)]^2} + c$
b
(b) Let $\log (\sec x + \tan x) = t \Rightarrow \sec x\,dx = dt$
Therefore $\int_{}^{} {\sec x\,\log (\sec x + \tan x)\,dx} = \int_{}^{} {t\,dt} $
$ = \frac{{{t^2}}}{2} + c = \frac{{{{[\log (\sec x + \tan x)]}^2}}}{2} + c.$

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