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

Summary Submit Skip Question

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,

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