int_0^{pi / 2} log |tan x+cot x| , dx= | vedclass

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(A) Let $I = \int_0^{\pi / 2} \log |	an x + \cot x| \, dx$. Since $	an x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x +

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$\pi \log 2$

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(A) Let $I = \int_0^{\pi / 2} \log |	an x + \cot x| \, dx$. Since $	an x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x} = \frac{2}{\sin 2x}$. Thus,$I = \int_0^{\pi / 2} \log |\frac{2}{\sin 2x}| \, dx = \int_0^{\pi / 2} (\log 2 - \log |\sin 2x|) \, dx$. $I = \int_0^{\pi / 2} \log 2 \, dx - \int_0^{\pi / 2} \log |\sin 2x| \, dx$. $I = \frac{\pi}{2} \log 2 - \int_0^{\pi / 2} \log |\sin 2x| \, dx$. Let $2x = t$,then $2 \, dx = dt$,so $dx = \frac{dt}{2}$. When $x=0, t=0$ and when $x=\frac{\pi}{2}, t=\pi$. $\int_0^{\pi / 2} \log |\sin 2x| \, dx = \frac{1}{2} \int_0^{\pi} \log |\sin t| \, dt = \frac{1}{2} 	imes 2 \int_0^{\pi / 2} \log |\sin t| \, dt = \int_0^{\pi / 2} \log |\sin t| \, dt$. Using the property $\int_0^{\pi / 2} \log |\sin t| \, dt = -\frac{\pi}{2} \log 2$. Therefore,$I = \frac{\pi}{2} \log 2 - (-\frac{\pi}{2} \log 2) = \frac{\pi}{2} \log 2 + \frac{\pi}{2} \log 2 = \pi \log 2$.

$\int_0^{\pi / 2} \log |\tan x+\cot x| \, dx=$

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