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Chapter 7: Problem 39

If \(y \in(1, \infty)\), then show that \(\sec ^{-1} y=\lim _{a \rightarrow 1^{+}} \int_{a}^{y} \frac{1}{t \sqrt{t^{2}-1}} d t \quad\) and \(\quad \csc ^{-1} y=\frac{\pi}{2}-\lim _{a \rightarrow 1^{+}} \int_{a}^{y} \frac{1}{t \sqrt{t^{2}-1}} d t .\)

Short Answer

Expert verified
The inverse secant function can be expressed as \(\sec^{-1} y = \lim_{a \rightarrow 1^+} \int_{a}^y \frac{1}{t \sqrt{t^2 - 1}} dt\), and the inverse cosecant function can be expressed as \(\csc^{-1} y = \frac{\pi}{2} - \lim_{a \rightarrow 1^+} \int_{a}^y \frac{1}{t \sqrt{t^2 - 1}} dt\). These expressions are derived using the antiderivatives of the derivatives of the respective inverse functions, along with their interrelationship.

Step by step solution

01

Expressing the inverse secant function as an integral

Recall that if \(y = \sec x\), then we can write \(x\) as an integral of the form: \(x = \int \frac{1}{y \sqrt{y^2 - 1}} dy\) By the fundamental theorem of calculus, we can evaluate the derivative of the inverse secant function as: \(\frac{d}{dy}(\sec^{-1} y) = \frac{1}{y \sqrt{y^2 - 1}}\) Now, we need to find an antiderivative of the function with respect to \(y\). We know that: \(\sec^{-1}y = \int_{a}^y \frac{1}{t \sqrt{t^2 - 1}} dt\) Where a is a constant value.
02

Expressing the inverse cosecant function as an integral

Similarly, if \(y = \csc x\), we can express \(x\) as an integral of the form: \(x = \int \frac{1}{y \sqrt{y^2 - 1}} dy\) We know that the relation between the cosecant function and the secant function is as follows: \(\csc x = \frac{\pi}{2} - \sec x\) So, we can express the inverse cosecant function in terms of the inverse secant function: \(\csc^{-1} y = \frac{\pi}{2} - \sec^{-1} y\) Next, we need to find an antiderivative of the derivative with respect to \(y\). We know that: \(\csc^{-1} y = \frac{\pi}{2} - \int_{a}^y \frac{1}{t \sqrt{t^2 - 1}} dt\)
03

Finding the limit expressions for the inverse secant and cosecant functions

Now, we will find the limit expressions for the inverse secant and inverse cosecant functions by taking limits as \(a \rightarrow 1^+\). For the inverse secant function, we get: \(\sec^{-1} y = \lim_{a \rightarrow 1^+} \int_{a}^y \frac{1}{t \sqrt{t^2 - 1}} dt\) For the inverse cosecant function, we get: \(\csc^{-1} y = \frac{\pi}{2} - \lim_{a \rightarrow 1^+} \int_{a}^y \frac{1}{t \sqrt{t^2 - 1}} dt\) Thus, we have demonstrated the required expressions for the inverse secant and inverse cosecant functions as given in the exercise.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inverse Trigonometric Functions
Inverse trigonometric functions are vital in calculus, often serving as the bridge from trigonometric ratios to corresponding angles. Specifically, the inverse secant, \(\sec^{-1} y\), and inverse cosecant, \(\csc^{-1} y\), functions help find angles when the hypotenuse and adjacent or opposite sides are known. These functions operate within specific intervals to maintain uniqueness of output.
  • Inverse Secant (\(\sec^{-1}\)): Takes as input a value greater than or equal to 1 or less than or equal to -1, producing an angle between \(0\) and \(\pi\), excluding \(\pi/2\).
  • Inverse Cosecant (\(\csc^{-1}\)): Accepts an input value less than or equal to -1 or greater than or equal to 1, yielding an angle in the ranges \( [ -\pi/2 , 0 ) \) and \( (0, \pi/2] \).

In our exercise, expressing these inverse functions as definite integrals facilitates their handling and differentiation, transforming complex trigonometric problems into manageable calculus tasks.
Definite Integrals
Definite integrals are a potent tool in calculus to evaluate the area under a curve between two points. Unlike indefinite integrals, which include a constant of integration, definite integrals provide an exact numerical value. They have a deep connection with the Fundamental Theorem of Calculus, which connects differentiation and integration.
  • In the exercise, we use definite integrals to express the inverse trigonometric functions. This allows us to compute exact values of the inverse functions through integration.
  • To evaluate definite integrals, a function must be integrable on the interval, which is often dependent on the nature of the function and the limits of integration.
  • Limits of integration, denoted here from \(a\) to \(y\), become crucial, particularly as we transition to limits as \(a ightarrow 1^+\).
The exercise demonstrates applying definite integrals to solve for \(\sec^{-1} y\) and \(\csc^{-1} y\), turning potential angles into manageable integral evaluations utilized with limits.
Limits
Limits are fundamental in calculus, providing a rigorous approach to understanding how functions behave as they approach specific points or infinity. In our exercise, limits handle behavior as \(a\) approaches \(1^+\), a point of significance.
  • Calculating limits involves examining what value a function approaches, rather than necessarily the function's value at that point.
  • This approach aids in defining functions' behavior at points of discontinuity or where immediate evaluation isn't feasible.
  • In the context of our exercise, as \(a ightarrow 1^+\), the integrals transform into exact expressions for inverse trigonometric functions. This action defines these functions precisely within necessary intervals.
Mastery of limits is crucial for understanding the deeper implications of calculus, especially when algebraic manipulation is insufficient or impossible for evaluating function behavior.

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Most popular questions from this chapter

Is it possible that $$ \ln x=\left(\sqrt[3]{e x^{2}+(\pi-2 e) x+e-\pi}+\sqrt{\pi x^{2}+(\sqrt{2}-2 \pi) x+\pi-\sqrt{2}}\right)^{1 / 17} $$ for all \(x>0\) ? Justify your answer.

Let \(r \in \mathbb{R}\) be positive and consider the function \(f:(0, \infty) \rightarrow \mathbb{R}\) defined by \(f(x)=x^{r}\). Show that the growth rate of \(\ln x\) is less than that of \(f\), while the growth rate of \(\exp x\) is more than that of \(f\) as \(x \rightarrow \infty\).

Prove the following for all \(y_{1}, y_{2} \in \mathbb{R}\) : (i) \(\tan ^{-1} y_{1}+\tan ^{-1} y_{2}=\tan ^{-1}\left(\frac{y_{1}+y_{2}}{1-y_{1} y_{2}}\right)\) if \(y_{1} y_{2}<1\), (ii) \(\tan ^{-1}\left|y_{1}\right|+\tan ^{-1}\left|y_{2}\right|=\frac{\pi}{2}\) if \(y_{1} y_{2}=1\), (iii) \(\tan ^{-1}\left|y_{1}\right|+\tan ^{-1}\left|y_{2}\right|=\tan ^{-1}\left(\frac{\left|y_{1}\right|+\left|y_{2}\right|}{1-y_{1} y_{2}}\right)\) if \(y_{1} y_{2}>1\).

Consider the function \(g: \mathbb{R} \backslash\\{0\\} \rightarrow \mathbb{R}\) defined by \(g(x):=\cos (1 / x)\). Prove the following: (i) \(g\) is an even function. (ii) \(\lim _{x \rightarrow 0} g(x)\) does not exist, but \(\lim _{x \rightarrow 0^{+}}(g(x)-g(-x))\) exists. Also, \(g\) cannot be extended to \(\mathbb{R}\) as a continuous function. (iii) For every \(\delta>0\), the function \(g\) is not uniformly continuous on \((0, \delta)\) as well as on \((-\delta, 0)\), but it is uniformly continuous on \((\infty,-\delta] \cup[\delta, \infty)\). (iv) For every \(\delta>0\), the function \(g\) is not monotonic, not convex, and not concave on \((0, \delta)\) as well as on \((-\delta, 0)\).

Consider the function \(h: \mathbb{R} \rightarrow \mathbb{R}\) defined by $$ h(x):=\left\\{\begin{array}{ll} |x|+|x \sin (1 / x)| & \text { if } x \neq 0 \\ 0 & \text { if } x=0 \end{array}\right. $$ Show that \(h\) has a strict absolute minimum at 0 , but for every \(\delta>0\), the function \(h\) is neither decreasing on \((-\delta, 0)\) nor increasing on \((0, \delta)\).
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