_0^ / 4 x ( x+ x) d x= - Vidyadip

MCQ

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

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

✓

Answer

Correct option: A.

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

(A)
Let $I =\int_0^{\pi / 4} \sec x \log (\sec x+\tan x) d x$
Put $\log (\sec x+\tan x)= t \Rightarrow \sec x d x= dt$
$\therefore \quad I =\int_0^{\log (\sqrt{2}+1)} tdt =\left[\frac{ t ^2}{2}\right]_0^{\log (\sqrt{2}+1)}=\frac{[\log (\sqrt{2}+1)]^2}{2}$

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