$int_{\frac{pi}{4}}^{\frac{pi}{2}}\frac{cos x}{sin x}dx=$

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 $x=\frac{\pi}{2}$, $u = \sin\frac{\pi}{2}=1$.

Step2: Rewrite the integral

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

Step3: Integrate

We know that $\int\frac{du}{u}=\ln|u|+C$. So $\int_{\frac{\sqrt{2}}{2}}^{1}\frac{du}{u}=\left[\ln u\right]_{\frac{\sqrt{2}}{2}}^{1}$.

Step4: Evaluate the definite - integral

$\ln(1)-\ln(\frac{\sqrt{2}}{2})=0 - (\ln\sqrt{2}-\ln2)=-\frac{1}{2}\ln2+\ln2=\frac{1}{2}\ln2$.

Answer:

$\frac{1}{2}\ln2$