If int_0^{frac{pi}{2}} log cos x , dx = frac{ | vedclass

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(D) We are given that $\int_0^{\frac{\pi}{2}} \log \cos x \, dx = \frac{\pi}{2} \log \left(\frac{1}{2}\right)$.We need to evaluate $I = \int_0^{\frac{

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$\frac{\pi}{2} \log 2$

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(D) We are given that $\int_0^{\frac{\pi}{2}} \log \cos x \, dx = \frac{\pi}{2} \log \left(\frac{1}{2}\right)$.We need to evaluate $I = \int_0^{\frac{\pi}{2}} \log \sec x \, dx$.Since $\sec x = \frac{1}{\cos x}$,we have $\log \sec x = \log \left(\frac{1}{\cos x}\right) = \log 1 - \log \cos x = -\log \cos x$.Therefore,$I = \int_0^{\frac{\pi}{2}} -\log \cos x \, dx$.$I = -\int_0^{\frac{\pi}{2}} \log \cos x \, dx$.Substituting the given value,$I = -\left[\frac{\pi}{2} \log \left(\frac{1}{2}\right)\right]$.Since $\log \left(\frac{1}{2}\right) = \log (2^{-1}) = -\log 2$,we get $I = -\left[\frac{\pi}{2} (-\log 2)\right] = \frac{\pi}{2} \log 2$.

If $\int_0^{\frac{\pi}{2}} \log \cos x \, dx = \frac{\pi}{2} \log \left(\frac{1}{2}\right)$,then $\int_0^{\frac{\pi}{2}} \log \sec x \, dx = $

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