The area bounded by $y = \sec^{- 1} x$ , $y = \cos e c^{- 1} x$ and line $x - 1 = 0$ is

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$π - \log_{e} 3 s q . u n i t s$

$\frac{π}{2} - \log (3 + 2 \sqrt{2}) s q . u n i t s$

$π - \log_{e} 3 \sqrt{2} s q . u n i t s$

$\log (3 + 2 \sqrt{2}) - \frac{π}{2} s q . u n i t s$

answer is A.

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Detailed Solution

$\begin{aligned} A = \int_{0}^{π / 4} \sec y d y + \int_{π / 4}^{π / 2} cosec y d y - 1 \times \frac{π}{2} \\ = [\ln (\sec y + \tan y)]_{0}^{π / 4} + {[\ln \tan \frac{y}{2}]}_{π / 4}^{π / 2} - \frac{π}{2} \\ = [\ln (\sqrt{2} + 1) - \ln 1] + \ln \tan \frac{π}{4} - \ln \tan \frac{π}{8} - \frac{π}{2} \\ = \ln (\sqrt{2} + 1) - \ln (\sqrt{2} - 1) - \frac{π}{2} \\ = \ln (\frac{\sqrt{2} + 1}{\sqrt{2} - 1}) - \frac{π}{2} \\ = \ln (3 + 2 \sqrt{2}) - \frac{π}{2} \end{aligned}$