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Question: \(\int _ { 0 } ^ { \pi / 4 } \sec x \log ( \sec x + \tan x ) d x =\)...

∫0π/4sec⁡xlog⁡(sec⁡x+tan⁡x)dx=\int _ { 0 } ^ { \pi / 4 } \sec x \log ( \sec x + \tan x ) d x =

A

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

B

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

C

12[log⁡(2−1)]2\frac { 1 } { 2 } [ \log ( \sqrt { 2 } - 1 ) ] ^ { 2 }

D

[log⁡(2−1)]2[ \log ( \sqrt { 2 } - 1 ) ] ^ { 2 }

Answer

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

Explanation

Solution

I=∫0π/4sec⁡xlog⁡(sec⁡x+tan⁡x)dxI = \int _ { 0 } ^ { \pi / 4 } \sec x \log ( \sec x + \tan x ) d x

Put log⁡(sec⁡x+tan⁡x)=t⇒sec⁡xdx=dt\log ( \sec x + \tan x ) = t \Rightarrow \sec x d x = d t

⇒I=∫0log⁡(2+1)tdt=[t22]0log⁡(2+1)=[log⁡(2+1)]22\Rightarrow I = \int _ { 0 } ^ { \log ( \sqrt { 2 } + 1 ) } t d t = \left[ \frac { t ^ { 2 } } { 2 } \right] _ { 0 } ^ { \log ( \sqrt { 2 } + 1 ) } = \frac { [ \log ( \sqrt { 2 } + 1 ) ] ^ { 2 } } { 2 }.