int_{frac{-3}{4}}^{frac{pi-6}{8}} log (sin (4 | vedclass

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Question

(B) Let $I = \int_{\frac{-3}{4}}^{\frac{\pi-6}{8}} \log (\sin (4 x+3)) \, dx$. Substitute $t = 4x + 3$,then $dt = 4 \, dx$,which implies $dx = \frac{d

Vedclass · Accepted Answer

$-\frac{\pi}{8} \log 2$

Vedclass · Answer

(B) Let $I = \int_{\frac{-3}{4}}^{\frac{\pi-6}{8}} \log (\sin (4 x+3)) \, dx$. Substitute $t = 4x + 3$,then $dt = 4 \, dx$,which implies $dx = \frac{dt}{4}$. When $x = \frac{-3}{4}$,$t = 4(\frac{-3}{4}) + 3 = 0$. When $x = \frac{\pi-6}{8}$,$t = 4(\frac{\pi-6}{8}) + 3 = \frac{\pi-6}{2} + 3 = \frac{\pi}{2} - 3 + 3 = \frac{\pi}{2}$. Thus,the integral becomes $I = \int_{0}^{\frac{\pi}{2}} \log (\sin t) \frac{dt}{4} = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \log (\sin t) \, dt$. Using the standard definite integral property $\int_{0}^{\frac{\pi}{2}} \log (\sin t) \, dt = -\frac{\pi}{2} \log 2$,we get: $I = \frac{1}{4} 	imes (-\frac{\pi}{2} \log 2) = -\frac{\pi}{8} \log 2$.

$\int_{\frac{-3}{4}}^{\frac{\pi-6}{8}} \log (\sin (4 x+3)) \, dx =$

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