Problem 91 - Calculus Early Transcendentals

Question

Evaluate the following integrals. \(\int_{5 / 12}^{3 / 4} \frac{\sinh ^{-1} x}{\sqrt{x^{2}+1}} d x\)

Step-by-Step Solution

Answer

Question: Evaluate the definite integral $$\int_{5/12}^{3/4} \frac{\sinh^{-1}(x)}{\sqrt{x^2 + 1}} dx$$. Answer: The value of the integral is $$\frac{1}{2}(\sinh^{-1}(3/4))^2 - \frac{1}{2}(\sinh^{-1}(5/12))^2$$.

Step 1: Perform substitution

Let \(u = \sinh^{-1}(x)\). Then \(x = \sinh(u)\). Now, we must find \(dx\) in terms of \(du\). To do this, we differentiate \(x\) with respect to \(u\): $$\frac{dx}{du} = \frac{d}{du} (\sinh(u)) = \cosh(u)$$ Now, we can get \(dx\) in terms of \(du\): $$dx = \cosh(u) du$$

Step 2: Change the limits of integration

Since we're changing the variable of integration from \(x\) to \(u\), we also need to change the limits of integration accordingly. Our new limits are: For the lower limit, when \(x = 5/12\): $$u = \sinh^{-1}(5/12)$$ For the upper limit, when \(x = 3/4\): $$u = \sinh^{-1}(3/4)$$

Step 3: Substitute and simplify the integrand

Substituting \(u\), \(x\), and \(dx\) into the integral gives: $$\int_{\sinh^{-1}(5/12)}^{\sinh^{-1}(3/4)} \frac{u}{\sqrt{\sinh^2(u) + 1}} \cosh(u) du$$ Recall the identity \(\cosh^2(u) - \sinh^2(u) = 1\). This can be used to simplify the integrand: $$\int_{\sinh^{-1}(5/12)}^{\sinh^{-1}(3/4)} \frac{u \cosh(u)}{\sqrt{\cosh^2(u)}} du$$ Since the square root of \(\cosh^2(u)\) is simply \(\cosh(u)\), the integrand simplifies to just \(u\): $$\int_{\sinh^{-1}(5/12)}^{\sinh^{-1}(3/4)} u du$$

Step 4: Evaluate the integral

The antiderivative of \(u\) is \(\frac{1}{2}u^2\), so the definite integral is: $$\left[\frac{1}{2}u^2\right]_{\sinh^{-1}(5/12)}^{\sinh^{-1}(3/4)} = \frac{1}{2}(\sinh^{-1}(3/4))^2 - \frac{1}{2}(\sinh^{-1}(5/12))^2$$ Thus, the value of the integral is: $$\int_{5/12}^{3/4} \frac{\sinh^{-1}(x)}{\sqrt{x^2 + 1}} dx = \frac{1}{2}(\sinh^{-1}(3/4))^2 - \frac{1}{2}(\sinh^{-1}(5/12))^2$$

Key Concepts

Inverse Hyperbolic Functions

Inverse hyperbolic functions are the inverses of the hyperbolic functions, much like how inverse trigonometric functions relate to their direct counterparts. The notation used is \( \sinh^{-1}(x) \), \( \cosh^{-1}(x) \), and so on. These functions allow us to find the value such that the hyperbolic function yields the given input. In this particular exercise, the inverse hyperbolic sine function, \( \sinh^{-1}(x) \), is crucial.
\( \sinh^{-1}(x) \) is defined as \( \ln(x + \sqrt{x^2 + 1}) \), reflecting the relation between hyperbolic functions and logarithms. Understanding this relationship is important because it helps simplify integration tasks that involve inverse hyperbolic functions. By translating inverse hyperbolic functions into logarithmic forms, we often make the integration process more manageable.
When working with integration involving inverse hyperbolic functions, remember:

• Understand their definitions and relationships with traditional logarithmic identities.
• Use their derivatives in integration by substitution, as was done in this exercise, to simplify the process.

Definite Integrals

Definite integrals represent the computation of the area under a curve from one point to another. In mathematical symbols, it is expressed as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration. In the given exercise, we computed the definite integral between the limits \( \frac{5}{12} \) and \( \frac{3}{4} \).
Definite integrals have limits which indicate the interval over which the area is calculated. Unlike indefinite integrals, which yield a family of functions plus a constant \( C \), definite integrals return a specific numerical value. To solve the definite integral:

• Substitute the limits of integration accurately after solving the indefinite integral.
• Ensure clarity in changing limits when substitution is involved.

Recall that in this exercise, once substitution was performed, it required changing from \( x \) to \( u = \sinh^{-1}(x) \), necessitating an adjustment of our integration limits.

Integration by Substitution

Integration by substitution is a powerful technique used to simplify complex integrals. It's akin to reverse chain rule in differentiation. In this exercise, we substituted \( u = \sinh^{-1}(x) \), transforming the original integral. This substitution helps convert the integrals into a more familiar and solvable form. Here’s how you can approach such problems:

• Identify a part of the integrand that can be substituted with a simpler function.
• Change both the integrand and the differential \( dx \) to terms of the new variable \( du \).
• Don't forget to change the limits of integration if the integral is definite.

This approach focuses on making the integration feasible by transforming it into a form where standard techniques apply. Substitution often involves differential change; remember, when function substitutions are made, the differential must also change to reflect the new variable. In our solution, this was reflected by translating \( x = \sinh(u) \) and thus \( dx = \cosh(u) du \), facilitating the evaluation of the definite integral efficiently.