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(A) Let $I = \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \ln(e^{\cos x}) \, d(\sin x)$.Since $\ln(e^{\cos x}) = \cos x$,the integral becomes $I = \int_{\f

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$\frac{\pi}{12}$

Vedclass · Answer

(A) Let $I = \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \ln(e^{\cos x}) \, d(\sin x)$.Since $\ln(e^{\cos x}) = \cos x$,the integral becomes $I = \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \cos x \, d(\sin x)$.We know that $d(\sin x) = \cos x \, dx$. When $\sin x = \frac{1}{2}$,$x = \frac{\pi}{6}$.When $\sin x = \frac{\sqrt{3}}{2}$,$x = \frac{\pi}{3}$.Substituting these,we get $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos x (\cos x \, dx) = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos^2 x \, dx$.Using the identity $\cos^2 x = \frac{1 + \cos 2x}{2}$,we have:$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1 + \cos 2x}{2} \, dx = \frac{1}{2} [x + \frac{\sin 2x}{2}]_{\frac{\pi}{6}}^{\frac{\pi}{3}}$.$I = \frac{1}{2} [(\frac{\pi}{3} + \frac{\sin(2\pi/3)}{2}) - (\frac{\pi}{6} + \frac{\sin(\pi/3)}{2})]$.$I = \frac{1}{2} [(\frac{\pi}{3} + \frac{\sqrt{3}}{4}) - (\frac{\pi}{6} + \frac{\sqrt{3}}{4})] = \frac{1}{2} [\frac{\pi}{3} - \frac{\pi}{6}] = \frac{1}{2} [\frac{\pi}{6}] = \frac{\pi}{12}$.

For $0 < x < \frac{\pi}{2}$,the integral $\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \ln(e^{\cos x}) \, d(\sin x)$ is equal to:

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