If I = intlimits_0^{frac{pi}{2}} ln(sin x) dx | vedclass

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(A) Let $I_1 = \int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\sin x + \cos x) dx$. Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we

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$\frac{I}{2}$

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(A) Let $I_1 = \int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\sin x + \cos x) dx$. Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we have: $I_1 = \int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\sin(-\frac{\pi}{4} + \frac{\pi}{4} - x) + \cos(-\frac{\pi}{4} + \frac{\pi}{4} - x)) dx = \int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\sin(-x) + \cos(-x)) dx = \int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\cos x - \sin x) dx$. Adding the two expressions for $I_1$: $2I_1 = \int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (\ln(\sin x + \cos x) + \ln(\cos x - \sin x)) dx = \int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\cos^2 x - \sin^2 x) dx = \int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\cos 2x) dx$. Since $\ln(\cos 2x)$ is an even function,$2I_1 = 2 \int\limits_0^{\frac{\pi}{4}} \ln(\cos 2x) dx$. Let $2x = t$,then $2dx = dt$. When $x=0, t=0$; when $x=\frac{\pi}{4}, t=\frac{\pi}{2}$. $I_1 = \int\limits_0^{\frac{\pi}{2}} \ln(\cos t) \frac{dt}{2} = \frac{1}{2} \int\limits_0^{\frac{\pi}{2}} \ln(\sin t) dt = \frac{I}{2}$.

If $I = \int\limits_0^{\frac{\pi}{2}} \ln(\sin x) dx$,then $\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\sin x + \cos x) dx =$

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