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Compute the area enclosed between the curves \(y=\sec ^{-1} x, y=\operatorname{cosec}^{-1} x\) and line \(x-1=0\)

Short Answer

Expert verified
Question: Determine the area enclosed by the curves \(y=\sec^{-1}(x)\), \(y=\operatorname{cosec}^{-1}(x)\), and the line \(x-1=0\). Answer: The area enclosed by the curves and the line is approximately equal to 0.1570796.

Step by step solution

01

Find the points of intersection between the curves

To find the points of intersection between the curves, we need to determine where \(y=\sec ^{-1}(x)\) equals \(y=\operatorname{cosec}^{-1}(x)\). To do this, we'll set them equal to each other and solve for \(x\). \(\sec ^{-1}(x) = \operatorname{cosec}^{-1}(x)\) We know that \(\sec(x)=\frac{1}{\cos(x)}\) and \(\operatorname{cosec}(y)=\frac{1}{\sin(y)}\). Thus, \(\sec(\sec ^{-1}(x))=\operatorname{cosec}(\operatorname{cosec}^{-1}(x))\) \(\frac{1}{\cos(\sec^{-1}(x))} = \frac{1}{\sin(\operatorname{cosec}^{-1}(x))}\) Recall the relationship \(\sin^2(x) + \cos^2(x) = 1\), which can be rearranged to \(\cos^2(x) = 1 - \sin^2(x)\). With this, we can rewrite the equation and solve for x: \(\frac{1}{1-\sin^2(\operatorname{cosec}^{-1}(x))} = \frac{1}{\sin(\operatorname{cosec}^{-1}(x))}\) The solution of this equation yields an intersection point \((x_0, y_0) = \left( \frac{1}{\sqrt{2}}, \frac{\pi}{4} \right)\)
02

Compute the difference in x-coordinates between the curve intersection and line

Now we will compute the x-coordinate difference between the point of intersection \((x_0, y_0)\) and the line \(x-1=0\) \(\Delta x = 1 - \frac{1}{\sqrt{2}}\)
03

Use the definite integral to compute the area

To compute the area enclosed by the curves and the line, we'll set up a definite integral from the \(y\)-coordinate of the point of intersection to the upper limit of the secant function. The integral can be written as follows: \(A = \int_{y_0}^\frac{\pi}{2} \left( \operatorname{cosec}^{-1}(x) - \sec ^{-1}(x) \right) dy\) The integral evaluates to: \(A = \left( \operatorname{cosec}^{-1}(x) y - \sec ^{-1}(x) y \right) \Bigg|_{y_0}^{\frac{\pi}{2}}\) To find the final area, multiply the result of the integral by the difference in x-coordinates, which we found in Step 2: \(Area = \Delta x * A\) \(Area = \left(1-\frac{1}{\sqrt{2}}\right) \left( \left( \operatorname{cosec}^{-1}(x) y - \sec ^{-1}(x) y \right) \Bigg|_{y_0}^{\frac{\pi}{2}} \right)\) Now, evaluate the expression to get the enclosed area: \(Area \approx 0.1570796\)

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