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Chapter 7: Problem 20

Evaluate each integral. \(\int \frac{\sin \sqrt{x}}{\sqrt{x}} d x\)

Short Answer

Expert verified
The integral evaluates to \(-2\cos(\sqrt{x}) + C\).

Step by step solution

01

Identify the Substitution

To simplify the integral, we need a substitution. Notice that the sine function's argument, \( \sqrt{x} \), suggests using a substitution related to \( \sqrt{x} \). Let's set \( u = \sqrt{x} \). Differentiating both sides, we get \( du = \frac{1}{2\sqrt{x}} \, dx \). Rearranging gives \( dx = 2u \, du \).
02

Substitute and Simplify the Integral

Substitute \( u = \sqrt{x} \) and \( dx = 2u \, du \) into the integral: \[\int \frac{\sin \sqrt{x}}{\sqrt{x}} \, dx = \int \frac{\sin u}{u} (2u) \, du = 2 \int \sin u \, du\] Now, the integral is simplified to \( 2 \int \sin u \, du \).
03

Integrate with Respect to \( u \)

Now, integrate \( 2 \int \sin u \, du \):\[2 \int \sin u \, du = 2(-\cos u) + C = -2\cos u + C\]We have integrated \( \sin u \) to get \( -\cos u \), then multiplied by 2.
04

Back-Substitute to Original Variable

Replace \( u \) with \( \sqrt{x} \) in the result of the integration:\[-2\cos u + C = -2\cos(\sqrt{x}) + C\]This is the solution to the original integral.

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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 fundamental tool in calculus that simplifies complex integrals into more manageable ones. It is particularly useful when dealing with composite functions such as trigonometric, exponential, or logarithmic functions.
  • When using substitution, we identify a part of the integral that can be replaced with a single variable.
  • In the context of the exercise, we substitute the inner function, which is \( u = \sqrt{x} \).
  • This changes the original differential \( dx \) into a simpler form, \( dx = 2u \, du \), making the integral much easier to solve.
Remember, the goal is to rewrite the integral in terms of \( u \) and \( du \) to make the integration process straightforward. If we choose the correct substitution, integrating becomes similar to a basic integration problem.
Definite Integrals
While this exercise focuses on indefinite integrals, definite integrals involve finding the area under a curve between two specific limits. Despite their differences, understanding indefinite integrals is crucial for mastering definite integrals.
  • A definite integral has the form \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the integration limits.
  • The substitution method can still be applied, but one must also adjust the limits of integration according to the substitution.
For example, if \( u = \sqrt{x} \), the limits, originally \( x \)-based, must be transformed to \( u \)-based. This ensures that when you compute the area, you're accurately representing the transformed function鈥檚 geometry.
Trigonometric Integrals
Trigonometric integrals are a specific class of integrals involving trigonometric functions like sine, cosine, and tangent. They appear frequently in calculus problems due to their periodic nature and broad applicability.
  • The integral in the exercise is a trigonometric integral because of the presence of the \( \sin(u) \) function.
  • Recognizing trigonometric functions allows us to apply known integration formulas such as \( \int \sin u \, du = -\cos u + C \).
  • Understanding these integrals enables us to find areas, solve differential equations, and model wave functions.
In calculus, mastering these integrals equips you with powerful tools for analyzing complex physical phenomena and functions with periodic behavior.

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