The area enclosed by the curves y = sin^{-1}( | vedclass

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(A) Given $y_1 = \sin^{-1}(\cos x) = \sin^{-1}(\sin(\frac{\pi}{2} - x))$. For $x \in [\frac{\pi}{2}, \frac{3\pi}{2}]$,$\frac{\pi}{2} - x \in [-\pi, 0]

Vedclass · Accepted Answer

$\frac{\pi^2}{4}$

Vedclass · Answer

(A) Given $y_1 = \sin^{-1}(\cos x) = \sin^{-1}(\sin(\frac{\pi}{2} - x))$. For $x \in [\frac{\pi}{2}, \frac{3\pi}{2}]$,$\frac{\pi}{2} - x \in [-\pi, 0]$. Thus,$y_1 = -x - \frac{\pi}{2}$ (for $x \in [\frac{\pi}{2}, \pi]$) and $y_1 = x - \frac{3\pi}{2}$ (for $x \in [\pi, \frac{3\pi}{2}]$).Given $y_2 = \cos^{-1}(\sin x) = \cos^{-1}(\cos(\frac{\pi}{2} - x))$. For $x \in [\frac{\pi}{2}, \frac{3\pi}{2}]$,$\frac{\pi}{2} - x \in [-\pi, 0]$. Thus,$y_2 = |\frac{\pi}{2} - x| = x - \frac{\pi}{2}$ (for $x \in [\frac{\pi}{2}, \pi]$) and $y_2 = -(\frac{\pi}{2} - x) = x - \frac{\pi}{2}$ is not correct,actually $y_2 = \frac{3\pi}{2} - x$ for $x \in [\pi, \frac{3\pi}{2}]$.The area is $\int_{\pi/2}^{\pi} (x - \frac{\pi}{2} - (\frac{\pi}{2} - x)) dx + \int_{\pi}^{3\pi/2} (\frac{3\pi}{2} - x - (x - \frac{3\pi}{2})) dx$ is incorrect based on the graph.Looking at the graph,the area is composed of two triangles. The first triangle has vertices $(\frac{\pi}{2}, 0), (\pi, \frac{\pi}{2}), (\pi, 0)$ and the second has vertices $(\pi, 0), (\pi, \frac{\pi}{2}), (\frac{3\pi}{2}, 0)$.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \frac{\pi}{2} 	imes \frac{\pi}{2} + \frac{1}{2} 	imes \frac{\pi}{2} 	imes \frac{\pi}{2} = \frac{\pi^2}{8} + \frac{\pi^2}{8} = \frac{\pi^2}{4}$.

The area enclosed by the curves $y = \sin^{-1}(\cos x)$ and $y = \cos^{-1}(\sin x)$ for $x \in \left[ \frac{\pi}{2}, \frac{3\pi}{2} \right]$ is:

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