_0^ /4 x ( x + x) ,dx = - Vidyadip

MCQ

$\int_0^{\pi /4} {} \sec x\log (\sec x + \tan x)\,dx = $

✓
$\frac{1}{2}{[\log (1 + \sqrt 2 )]^2}$
B
${[\log (1 + \sqrt 2 )]^2}$
C
$\frac{1}{2}{[\log (\sqrt 2 - 1)]^3}$
D
$\frac{1}{2}{[\log (\sqrt 2 - 1)]^2}$

✓

Answer

Correct option: A.

$\frac{1}{2}{[\log (1 + \sqrt 2 )]^2}$

a
(a) $I = \int_0^{\pi /4} {\sec x\log (\sec x + \tan x)dx} $

Put $\log (\sec x + \tan x) = t \Rightarrow \sec x\,dx = dt$

$ \Rightarrow I = \int_0^{\log (\sqrt 2 + 1)} {t\,dt = \left[ {\frac{{{t^2}}}{2}} \right]} _0^{\log (\sqrt 2 + 1)} $

$= \frac{{{{[\log (\sqrt 2 + 1)]}^2}}}{2}$.

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