Evaluate the definite integral: int_0^{pi /2} | vedclass

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(D) Let $I = \int_0^{\pi /2} \log(	an x) \, dx$. Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get: $I = \int_0^{\pi /2} \log(\

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$0$

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(D) Let $I = \int_0^{\pi /2} \log(	an x) \, dx$. Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get: $I = \int_0^{\pi /2} \log(	an(\frac{\pi}{2} - x)) \, dx$ $I = \int_0^{\pi /2} \log(\cot x) \, dx$ $I = \int_0^{\pi /2} \log(\frac{1}{	an x}) \, dx$ $I = \int_0^{\pi /2} (\log 1 - \log(	an x)) \, dx$ Since $\log 1 = 0$,we have: $I = - \int_0^{\pi /2} \log(	an x) \, dx$ $I = -I$ $2I = 0 \implies I = 0$.

Evaluate the definite integral: $\int_0^{\pi /2} \log(\tan x) \, dx$.

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