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

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(D) આપણને આપેલ છે કે $\int_0^{\frac{\pi}{2}} \log \cos x \, dx = \frac{\pi}{2} \log \left(\frac{1}{2}\right)$.આપણે $I = \int_0^{\frac{\pi}{2}} \log \s

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) આપણને આપેલ છે કે $\int_0^{\frac{\pi}{2}} \log \cos x \, dx = \frac{\pi}{2} \log \left(\frac{1}{2}\right)$.આપણે $I = \int_0^{\frac{\pi}{2}} \log \sec x \, dx$ ની કિંમત શોધવાની છે.કારણ કે $\sec x = \frac{1}{\cos x}$,તેથી $\log \sec x = \log \left(\frac{1}{\cos x}\right) = \log 1 - \log \cos x = -\log \cos x$ થાય.તેથી,$I = \int_0^{\frac{\pi}{2}} -\log \cos x \, dx$.$I = -\int_0^{\frac{\pi}{2}} \log \cos x \, dx$.આપેલ કિંમત મૂકતા,$I = -\left[\frac{\pi}{2} \log \left(\frac{1}{2}\right)\right]$.કારણ કે $\log \left(\frac{1}{2}\right) = \log (2^{-1}) = -\log 2$,તેથી $I = -\left[\frac{\pi}{2} (-\log 2)\right] = \frac{\pi}{2} \log 2$ મળે.

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

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