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

Explanation:

Step1: Use substitution

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 $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx$ becomes $\int_{\frac{\sqrt{2}}{2}}^{1}\frac{du}{u}$.

Step2: Integrate $\frac{1}{u}$

The antiderivative of $\frac{1}{u}$ is $\ln|u|$. So $\int_{\frac{\sqrt{2}}{2}}^{1}\frac{du}{u}=\left[\ln u\right]_{\frac{\sqrt{2}}{2}}^{1}$.

Step3: Evaluate the definite - integral

$\left[\ln u\right]{\frac{\sqrt{2}}{2}}^{1}=\ln1-\ln\frac{\sqrt{2}}{2}=0 - (\ln\sqrt{2}-\ln2)=\ln2-\frac{1}{2}\ln2=\frac{1}{2}\ln2=\ln\sqrt{2}=\ln\sqrt{3}-\ln\sqrt{\frac{3}{2}}$. And $\int{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\ln\frac{1}{\frac{\sqrt{2}}{2}}=\ln\sqrt{2}=\ln\sqrt{3}-\ln\sqrt{\frac{3}{2}}=\ln\sqrt{3}$.

Answer:

A. $\ln\sqrt{3}$