/*! 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 31 Evaluate the integrals using app... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 31

Evaluate the integrals using appropriate substitutions. $$\int e^{\sin x} \cos x d x$$

Short Answer

Expert verified
\(e^{\sin x} + C\)

Step by step solution

01

Identify a Suitable Substitution

The integral \(\int e^{\sin x} \cos x \, dx\) suggests the substitution method because the derivative of \(\sin x\) is \(\cos x\). This matches a part of our integrand. Let \(u = \sin x\). Then \(du = \cos x \, dx\).
02

Substitute in the Integral

Substitute \(u = \sin x\) and \(du = \cos x \, dx\) into the integral. This gives us \(\int e^u \, du\).
03

Integrate with Respect to \(u\)

The integral \(\int e^u \, du\) is straightforward; the antiderivative of \(e^u\) is \(e^u\). Thus, \(\int e^u \, du = e^u + C\), where \(C\) is the constant of integration.
04

Back-Substitute to Original Variable

Replace \(u\) with \(\sin x\) to express the solution in terms of \(x\): \(e^u + C = e^{\sin x} + C\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Substitution Method
The substitution method is a powerful technique used in calculus to simplify complex integrals. It involves replacing a part of the integrand with a new variable to make the integral easier to evaluate. When identifying suitable substitutions, it's best to look for a function and its derivative in the integrand.

Here's a simple approach to applying substitution:
  • Identify a function inside the integrand whose derivative is also present as a factor. In our example, \( \sin x \) is chosen because its derivative, \( \cos x \), is also part of the integrand.
  • Replace this function with a new variable \( u \). For example, set \( u = \sin x \).
  • Calculate \( du = \cos x \ dx \) to find the differential of the new variable.
  • Substitute \( u \) and \( du \) into the integral and solve the resultant simpler integral.
This method reduces complex expressions into manageable ones, aiding the integration process.
Definite Integrals
Definite integrals help calculate the area under a curve within a specific interval [a, b]. Unlike indefinite integrals, which include a constant of integration, definite integrals provide a numerical result representing this area. When using the substitution method for definite integrals, an additional step is required to adjust the limits of integration.

Here's what you need to remember:
  • Determine the new limits of integration by substituting the original limits into the expression for \( u \). If \( x = a \,\ u = f(a) \) and if \( x = b \,\ u = f(b) \).
  • Evaluate the integral using the new variable and its limits to find the exact area.
  • The result of a definite integral is a number, indicating the net area considering regions above and below the x-axis as positive and negative, respectively.
By understanding these steps, one can seamlessly transition between solving indefinite and definite integrals using the substitution method.
Antiderivatives
An antiderivative, also known as the indefinite integral, is the reverse process of differentiation. It involves finding a function whose derivative results in the original integrand. When integrating functions, identifying their antiderivatives is crucial. In our problem, recognizing that the antiderivative of \( e^u \) is \( e^u \) simplifies the process immensely.

To master finding antiderivatives, focus on these aspects:
  • Recognize common functions and their antiderivatives, like polynomials, exponentials, and trigonometric functions.
  • Use the fact that the antiderivative of \( e^u \) is \( e^u + C \,\) where \ C \ is the constant of integration.
  • If a substitution has been made, remember to return to the original variable by back-substituting once the integration is complete.
Learning antiderivatives unlocks the ability to solve a variety of integrals, both simple and complex. Practicing with a range of functions is the key to developing proficiency in this essential area of calculus.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) Give a geometric argument to show that \(\frac{1}{x+1}<\int_{x}^{x+1} \frac{1}{t} d t<\frac{1}{x}, \quad x>0\) (b) Use the result in part (a) to prove that $$ \frac{1}{x+1}<\ln \left(1+\frac{1}{x}\right)<\frac{1}{x}, \quad x>0 $$ (c) Use the result in part (b) to prove that $$ e^{x /(x+1)}<\left(1+\frac{1}{x}\right)^{x}0 $$ and hence that $$ \lim _{x \rightarrow+\infty}\left(1+\frac{1}{x}\right)^{x}=e $$ (d) Use the result in part (b) to prove that $$ \left(1+\frac{1}{x}\right)^{x}0 $$

It can be shown that every interval contains both rational and irrational numbers. Accepting this to be so, do you believe that the function $$ f(x)=\left\\{\begin{array}{lll} 1 & \text { if } & x \text { is rational } \\ 0 & \text { if } & x \text { is irrational } \end{array}\right. $$ is integrable on a closed interval \([a, b] ?\) Explain your reasoning.

Evaluate the integrals using appropriate substitutions. $$\int \frac{y}{\sqrt{2 y+1}} d y$$

Sketch the region described and find its area. The region under the curve \(y=3 \sin x\) and over the interval \([0,2 \pi / 3]\)

$$\begin{aligned} \sum_{k=1}^{4}\left[(k+1)^{3}-k^{3}\right]=&\left[5^{3}-4^{3}\right]+\left[4^{3}-3^{3}\right] \\\ &+\left[3^{3}-2^{3}\right]+\left[2^{3}-1^{3}\right] \\ =& 5^{3}-1^{3}=124 \end{aligned}$$ For convenience, the terms are listed in reverse order. Note how cancellation allows the entire sum to collapse like a telescope. A sum is said to telescope when part of each term cancels part of an adjacent term, leaving only portions of the first and last terms uncanceled. Evaluate the telescoping sums in these exercises. $$\sum_{k=2}^{20}\left(\frac{1}{k^{2}}-\frac{1}{(k-1)^{2}}\right)$$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.