/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 90 Evaluate the integrals in Exerci... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the integrals in Exercises \(83-92.\) $$\int_{0}^{\pi / 4}\left(\frac{1}{3}\right)^{\tan t} \sec ^{2} t d t$$

Short Answer

Expert verified
The integral evaluates to \( \frac{2}{3|\ln(3)|} \).

Step by step solution

01

Recognize the function and its derivative

The integrand is \( \left(\frac{1}{3}\right)^{\tan t} \sec^2 t \). Recognize that if we let \( u = \tan t \), then the derivative \( \frac{du}{dt} = \sec^2 t \). This means the structure of the integral suggests using a substitution method since \( \sec^2 t dt = du \).
02

Use substitution to change the variable

With the substitution \( u = \tan t \), the differential becomes \( du = \sec^2 t \, dt \). Substituting these into the integral changes it to \( \int \left(\frac{1}{3}\right)^{u} \, du \).
03

Evaluate the new integral

The integral \( \int \left(\frac{1}{3}\right)^u \, du \) is equal to \( \int e^{u \ln(1/3)} \, du \), which evaluates to \( \frac{e^{u \ln(1/3)}}{\ln(1/3)} + C \).
04

Change limits of integration

Convert the limits from \( t \) to \( u \). When \( t = 0 \), \( u = \tan 0 = 0 \). When \( t = \pi/4 \), \( u = \tan(\pi/4) = 1 \). Thus, the limits of integration in terms of \( u \) are from 0 to 1.
05

Substitute back and evaluate the definite integral

Substitute back to get \( \frac{(1/3)^u}{\ln(1/3)} \) and evaluate from 0 to 1:\[ \left[ \frac{(1/3)^u}{\ln(1/3)} \right]_0^1 = \frac{(1/3)^1 - (1/3)^0}{\ln(1/3)} = \frac{1/3 - 1}{\ln(1/3)} = \frac{-2/3}{\ln(1/3)}. \]
06

Compute the expression

Simplifying, we find: \( \frac{-2/3}{\ln(1/3)} \), which can be rewritten using the properties of logarithms as \( \frac{2}{3 |\ln(3)|} \) because \( \ln(1/3) = -\ln(3) \).

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Key Concepts

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

Substitution Method
The substitution method is a powerful tool in calculus for solving integrals. It's particularly useful when you encounter integrals with a function and its derivative present. The idea is to simplify the integral by substituting a part of it with a new variable. This method often makes a complex integral easier to evaluate.

In the exercise provided, we noticed that the integrand is \( \left(\frac{1}{3}\right)^{\tan t} \sec^2 t \). The presence of \( \tan t \) and its derivative \( \sec^2 t \) suggests substitution could be beneficial.
  • To apply the substitution, set a new variable such as \( u = \tan t \).
  • Differentiate it to find \( du = \sec^2 t \, dt \).
When substituted into the integral, it simplifies the variable and integrates in terms of \( u \), turning the understanding of a trigonometric function into a more straightforward exponential function problem in this particular case.
Definite Integrals
Definite integrals are used to compute the net area under a curve within given boundaries. They are defined with limits of integration, and the result is a specific numerical value, unlike indefinite integrals where we have a general function plus a constant.

For the integral in consideration, the bounds are from \( t = 0 \) to \( t = \pi/4 \). Upon substitution, these limits change in terms of our new variable \( u \), as follows:
  • When \( t = 0 \), \( u = \tan(0) = 0 \).
  • When \( t = \pi/4 \), \( u = \tan(\pi/4) = 1 \).
This transforms the integral into evaluating the area from 0 to 1 in the new variable. The process not only clarifies computations but also ensures we're working within the appropriate framework for solving the given problem. The final numerical result represents the accumulated change or area from the start to end within the specified range.
Trigonometric Functions
Trigonometric functions are fundamental in mathematics, often appearing in integrals and derivatives. They describe relationships between angles and sides of triangles and are periodic, meaning they repeat their values in regular intervals.

In the provided exercise, the function \( \tan t \) and its derivative \( \sec^2 t \) play a crucial role.
  • \( \tan t \) is the ratio of the opposite side to the adjacent side in a right triangle.
  • \( \sec^2 t \), the derivative of \( \tan t \), represents the rate of change of the tangent function, which is pivotal for substitution.
The integral's set-up relies on the interplay between \( \tan t \) and \( \sec^2 t \), using their mathematical properties to transform and solve the integral effectively. Understanding these functions can illuminate how different integrals can be approached and solved.

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Most popular questions from this chapter

To encourage buyers to place 100-unit orders, your firm's sales department applies a continuous discount that makes the unit price a function \(p(x)\) of the number of units \(x\) ordered. The discount decreases the price at the rate of \(\$ 0.01\) per unit ordered. The price per unit for a 100 -unit order is \(p(100)=\$ 20.09\) a. Find \(p(x)\) by solving the following initial value problem: $$\begin{aligned} &\text { Differential equation: } \quad \frac{d p}{d x}=-\frac{1}{100} p\\\ &\text { Initial condition: } \quad p(100)=20.09 \end{aligned}$$ b. Find the unit price \(p(10)\) for a 10 -unit order and the unit price \(p(90)\) for a 90 -unit order. c. The sales department has asked you to find out if it is discounting so much that the firm's revenue, \(r(x)=x \cdot p(x),\) will actually be less for a 100 -unit order than, say, for a 90 -unit order. Reassure them by showing that \(r\) has its maximum value at \(x=100\) d. Graph the revenue function \(r(x)=x p(x)\) for \(0 \leq x \leq 200\)

Evaluate the integrals in Exercises \(67-74\) in terms of a. inverse hyperbolic functions. b. natural logarithms. $$\int_{5 / 4}^{2} \frac{d x}{1-x^{2}}$$

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Find the area of the region between the curve \(y=2^{1-x}\) and the interval \(-1 \leq x \leq 1\) of the \(x\) -axis.

Use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$y=(\ln x)^{\ln x}$$

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