The area of the region bounded by the $x$-axis and the curves defined by $y= an x,-\frac{\pi}{3} \l

Question

The area of the region bounded by the $x$-axis and the curves defined by $y=	an x,-\frac{\pi}{3} \leq x \leq \frac{\pi}{3}$ $y=\cot x, \frac{\pi}{6} \leq x \leq \frac{\pi}{2}$ is ____sq units

Professor PC Thomas Classes · Answer

The curves intersect at $P$ where $	an x=\cot x$.\ i.e. $P=\left(\frac{\pi}{4}, 1ight)$ where $x=\pi / 4$ within the given domain of $x$.\ Required area contained in the given domain is the shaded portion shown in the figure.\ $A=\int_{\pi / 6}^{\pi / 4} 	an x d x+\int_{\pi / 4}^{\pi / 3} \cot x d x$\ $=[\log \sec x]_{\pi / 6}^{\pi / 4}+[\log \sin x]_{\pi / 4}^{\pi / 3}$\ $=\left(\log \sqrt{2}-\log \frac{2}{\sqrt{3}}ight)+\left(\log \frac{\sqrt{3}}{2}-\log \frac{1}{\sqrt{2}}ight)$\ $=\log \sqrt{2}+\log \frac{\sqrt{3}}{2}+\log \frac{\sqrt{3}}{2}+\log \sqrt{2}$\ $=2\left(\log \frac{\sqrt{2} \cdot \sqrt{3}}{2}ight)=2 \log \sqrt{\frac{3}{2}}=\log \frac{3}{2}$ sq. units\