$\displaystyle \int \limits_{\frac{\pi}{2}}^{\frac{\pi}{4}}-\frac{\cos{x}}{\sin{x}}   dx =$

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We can rewrite the given definite integral:

$$\int \limits_{\frac{\pi}{2}}^{\frac{\pi}{4}}-\frac{\cos{x}}{\sin{x}}   dx = \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos{x}}{\sin{x}}   dx$$

If we know the integration rule of trigonometric functions,

$$\int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos{x}}{\sin{x}}   dx = \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot{x}   dx$$ $$= \Big[ \ln(\sin{x}) \Big]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$$ $$= \ln\left(\sin \frac{\pi}{2}\right) - \ln\left(\sin\frac{\pi}{4}\right)$$ $$= \ln 1 - \ln \frac{\sqrt{2}}{2} = 0 - \ln\frac{\sqrt{2}}{2}$$ $$= \ln\left(\frac{\sqrt{2}}{2}\right)^{-1} = \ln\frac{2}{\sqrt{2}}$$ $$= \boxed{\ln\sqrt{2}}$$

By recognizing that the answer choices all contain $\ln$, we can try anticipating integration to $\ln$. With $\sin{x}$ in the denominator, we can anticipate:

$$f=\ln(\sin{x})$$

Using the derivative rule for the natural logarithmic function and chain rule, we arrive at our desired expression,

$$f'=\frac{\cos{x}}{\sin{x}}$$

Jump to Approach 1 for the corresponding steps.

We can use the substitution rule to integrate:

$$\int \limits_{\frac{\pi}{2}}^{\frac{\pi}{4}}-\frac{\cos{x}}{\sin{x}}   dx = \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos{x}}{\sin{x}}   dx$$

Setting $u=\sin{x}$

Solving for $dx$

$$u=\sin{x}$$ $$\frac{du}{dx}=\cos{x}$$ $$\frac{du}{\cos{x}}=dx$$

Therefore,

$$\int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos{x}}{\sin{x}}   dx = \int \limits_{\frac{\sqrt{2}}{2}}^{1} \frac{\cos{x}}{u} \cdot \frac{du}{\cos{x}} = \int \limits_{\frac{\sqrt{2}}{2}}^{1} \frac{1}{u}   du$$ $$= \Big[ \ln u \Big] _{\frac{\sqrt{2}}{2}}^{1}$$ $$= \ln 1 - \ln \frac{\sqrt{2}}{2} = 0 - \ln\frac{\sqrt{2}}{2}$$ $$= \ln\left(\frac{\sqrt{2}}{2}\right)^{-1} = \ln\frac{2}{\sqrt{2}}$$ $$= \boxed{\ln\sqrt{2}}$$