The area of the region bounded by the curve y | vedclass

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(C) The area $A$ is given by the integral of the function $y = \cot x$ from $x = \frac{\pi}{4}$ to $x = \frac{\pi}{2}$.$A = \int_{\frac{\pi}{4}}^{\fra

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$\frac{1}{2} \log 2$

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(C) The area $A$ is given by the integral of the function $y = \cot x$ from $x = \frac{\pi}{4}$ to $x = \frac{\pi}{2}$.$A = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot x \, dx$We know that $\int \cot x \, dx = \log |\sin x| + C$.Applying the limits:$A = [\log |\sin x|]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$$A = \log |\sin(\frac{\pi}{2})| - \log |\sin(\frac{\pi}{4})|$$A = \log(1) - \log(\frac{1}{\sqrt{2}})$Since $\log(1) = 0$ and $\log(\frac{1}{\sqrt{2}}) = \log(2^{-1/2}) = -\frac{1}{2} \log 2$:$A = 0 - (-\frac{1}{2} \log 2) = \frac{1}{2} \log 2$.Thus,the correct option is $C$.

The area of the region bounded by the curve $y = \cot x$,the lines $x = \frac{\pi}{4}$,$x = \frac{\pi}{2}$,and the $X$-axis is . . . . . . sq. units.

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