Textbook Question
Suppose that the function f and its derivative with respect to x have the following values at x=0, 1, 2, 3, and 4.
Assuming the inverse function f^(-1) is differentiable, find the slope of f^(-1)(x) at
c. x=3
Ch. 7 - Transcendental Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
Not the one you use?Change textbookSelect a different textbook
Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 7, Problem 7.7.73b
Evaluate the integrals in Exercises 67–74 in terms of
b. natural logarithms.
73. ∫(from 0 to π)cos(x)dx/√(1+sin²x)
Verified step by step guidance1
Identify the integral to evaluate: \(\displaystyle \int_0^{\pi} \frac{\cos(x)}{\sqrt{1 + \sin^2(x)}} \, dx\).
Recognize that the integrand contains \(\cos(x)\) and \(\sin(x)\), suggesting a substitution involving \(\sin(x)\) might simplify the integral.
Use the substitution \(u = \sin(x)\), which implies \(du = \cos(x) \, dx\). This transforms the integral into \(\int_{u=\sin(0)}^{u=\sin(\pi)} \frac{1}{\sqrt{1 + u^2}} \, du\).
Evaluate the new limits: since \(\sin(0) = 0\) and \(\sin(\pi) = 0\), the integral becomes \(\int_0^0 \frac{1}{\sqrt{1 + u^2}} \, du\).
Notice that the limits are the same, so the integral evaluates to zero. However, consider the behavior of the function on the interval and whether the substitution covers the entire integral correctly, possibly splitting the integral at \(\pi/2\) to handle the sign of \(\cos(x)\).
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
3mWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Definite Integrals
A definite integral calculates the net area under a curve between two specific limits. It is represented as ∫ from a to b of f(x) dx, where a and b are the bounds. Evaluating definite integrals often involves finding an antiderivative and then applying the Fundamental Theorem of Calculus.
Recommended video:
05:43
Definition of the Definite Integral
Substitution Method
The substitution method simplifies integrals by changing variables to transform the integral into a more manageable form. It is especially useful when the integrand contains a composite function, allowing the integral to be expressed in terms of a new variable, often leading to easier integration.
Recommended video:
07:33Euler's Method
Natural Logarithms in Integration
Natural logarithms often appear when integrating functions of the form f'(x)/f(x), resulting in ln|f(x)| + C. Recognizing when an integral can be expressed in terms of natural logarithms helps simplify the evaluation, especially when the integrand involves expressions under a root or rational functions.
Recommended video:
05:18Derivative of the Natural Logarithmic Function
Related Practice
Textbook Question
Use reference triangles in an appropriate quadrant to find the angles in Exercises 1–8.
4. b. arcsin(-1/√2)
Textbook Question
75. b. Identify the function’s local and absolute extreme values, if any, saying where they occur.
g(x) = x(ln x)²
Textbook Question
3. Use the properties of logarithms to write the expressions in Exercises 3 and 4 as a single term.
b. ln(3x² - 9x) + ln(1/3x)
Textbook Question
c. Find the slopes of the tangent lines to the graphs of h and k at (2, 2) and (−2, −2).
Textbook Question
In Exercises 67–72, you will explore some functions and their inverses together with their derivatives and tangent line approximations at specified points. Perform the following steps using your CAS:
c. Find the equation for the tangent line to f at the specified point (x_0, f(x_0)).
72. y= 2-x-x³, -2 ≤ x ≤ 2, x_0 = 3/2