$\\int_{\\frac{\\pi}{4}}^{\\frac{\\pi}{2}}\\frac{\\cos x}{\\sin x}dx=$\na $\\ln\\sqrt{3}$\nb…

Explanation:

Step1: Use substitution method

Let (u = \sin x), then (du=\cos xdx). When (x = \frac{\pi}{4}), (u=\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}); when (x=\frac{\pi}{2}), (u=\sin\frac{\pi}{2} = 1). The integral (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\int_{\frac{\sqrt{2}}{2}}^{1}\frac{du}{u}).

Step2: Calculate the integral

According to the integral formula (\int\frac{du}{u}=\ln|u|+C), we have (\int_{\frac{\sqrt{2}}{2}}^{1}\frac{du}{u}=\left[\ln u\right]{\frac{\sqrt{2}}{2}}^{1}). Substitute the upper - and lower - limits: (\ln1-\ln\frac{\sqrt{2}}{2}). Since (\ln1 = 0), and (\ln\frac{\sqrt{2}}{2}=\ln\sqrt{2}-\ln2=\frac{1}{2}\ln2-\ln2=-\frac{1}{2}\ln2). Also, (\ln\sqrt{3}=\frac{1}{2}\ln3\neq\ln\sqrt{\frac{3}{2}}=\frac{1}{2}(\ln3 - \ln2)), (\ln e = 1). Another way: (\int{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{d(\sin x)}{\sin x}=\left[\ln(\sin x)\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}) (=\ln(\sin\frac{\pi}{2})-\ln(\sin\frac{\pi}{4})=\ln1-\ln\frac{\sqrt{2}}{2}=-\left(\ln\sqrt{2}-\ln2\right)=-\left(\frac{1}{2}\ln2-\ln2\right)=\frac{1}{2}\ln2) (\ln\sqrt{3}=\frac{1}{2}\ln3\neq\frac{1}{2}\ln2), (\ln\sqrt{\frac{3}{2}}=\frac{1}{2}(\ln3 - \ln2)\neq\frac{1}{2}\ln2), (\ln e = 1\neq\frac{1}{2}\ln2)

Answer:

A. (\ln\sqrt{3})