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Magazine Chapter 5: Problem 52
If you do not know what substitution to make, try reducing the integral step by step, using a trial substitution to simplify the integral a bit and then another to simplify it some more. You will see what we mean if you try the sequences of substitutions in Exercises 51 and \(52 .\) $$ \int \sqrt{1+\sin ^{2}(x-1)} \sin (x-1) \cos (x-1) d x $$ $$ \begin{array}{l}{\text { a. } u=x-1, \text { followed by } v=\sin u, \text { then by } w=1+v^{2}} \\ {\text { b. } u=\sin (x-1) \text { , followed by } v=1+u^{2}} \\ {\text { c. } u=1+\sin ^{2}(x-1)}\end{array} $$
Short Answer
Expert verified
The integral evaluates to \( \frac{1}{3} (1 + \sin^2(x-1))^{3/2} + C \) for all approaches.
Step by step solution
01
Initial Substitution
Let us start by making the substitution for option (a): \( u = x - 1 \). Then, we have \( du = dx \), and the integral becomes \[ \int \sqrt{1 + \sin^2(u)} \sin(u) \cos(u) \, du. \]
02
Second Substitution for Option a
Continuing with the sequence (a), substitute \( v = \sin(u) \), which gives \( dv = \cos(u) \, du \). This changes the integral into \[ \int \sqrt{1 + v^2} \, v \, dv. \]
03
Third Substitution for Option a
Further simplifying using option (a), employ \( w = 1 + v^2 \). We have \( dw = 2v \, dv \) or \( v \, dv = \frac{1}{2} \, dw \). The integral then simplifies to \[ \frac{1}{2} \int \sqrt{w} \, dw. \]
04
Integration for Option a
Compute the integral: \[ \frac{1}{2} \int w^{1/2} \, dw = \frac{1}{2} \times \frac{2}{3} w^{3/2} + C = \frac{1}{3} w^{3/2} + C. \]
05
Back Substitution for Option a
Replace \( w \) with \( 1 + v^2 \), where \( v = \sin(u) \), and \( u = x - 1 \). The final answer is \[ \frac{1}{3} (1 + \sin^2(x - 1))^{3/2} + C. \]
06
Initial Substitution for Option b
Now try option (b) with the substitution \( u = \sin(x-1) \). Then, \( du = \cos(x-1) \, dx \), and the integral becomes \[ \int \sqrt{1 + u^2} \, u \, du. \]
07
Simplification for Option b
Perform the next substitution in option (b): Let \( v = 1 + u^2 \), giving \( dv = 2u \, du \), so \( u \, du = \frac{1}{2} \, dv \). The integral simplifies to \[ \frac{1}{2} \int \sqrt{v} \, dv. \]
08
Integration for Option b
Compute the integral: \[ \frac{1}{2} \int v^{1/2} \, dv = \frac{1}{2} \times \frac{2}{3} v^{3/2} = \frac{1}{3} v^{3/2} + C. \]
09
Back Substitution for Option b
Replace \( v \) with \( 1 + \sin^2(x-1) \). The final integral for this substitution is \[ \frac{1}{3} (1 + \sin^2(x-1))^{3/2} + C. \]
10
Substitution for Option c
For option (c), substitute \( u = 1 + \sin^2(x-1) \). This means the derivative is \( du = 2\sin(x-1)\cos(x-1) \, dx \), which leads to \[ dx = \frac{du}{2\sin(x-1)\cos(x-1)}. \]
11
Simplification for Option c
The integral changes to \[ \int \sqrt{u} \frac{1}{2} \, du = \frac{1}{2} \int u^{1/2} \, du. \]
12
Integration for Option c
Compute the integral: \[ \frac{1}{2} \int u^{1/2} \, du = \frac{1}{2} \times \frac{2}{3} u^{3/2} = \frac{1}{3} u^{3/2} + C. \]
13
Back Substitution for Option c
Replace \( u \) with \( 1 + \sin^2(x-1) \) to get the final answer: \[ \frac{1}{3} (1 + \sin^2(x-1))^{3/2} + C. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Trigonometric Integration
Trigonometric integration involves integrating expressions that include trigonometric functions, such as sine and cosine. These integrations can often look complex, but can be simplified using substitution techniques. In the exercise, we began by making suitable substitutions to simplify the integral step by step. For instance, by letting variables represent parts of the trigonometric function (e.g., \( u = x - 1 \) in the problem), the integral's complexity is reduced. This method requires identifying the part of the trigonometric expression that can be substituted to transform the integral into a simpler form.
Understanding trigonometric integration and its substitution methods is crucial for breaking down complex trigonometric expressions into easy-to-handle parts.
- Strategy: Replace trigonometric components with simpler variables to make the equation easier to integrate.
- Example: Transform \( \sqrt{1 + \sin^2(x-1)} \sin(x-1) \cos(x-1) \) to simpler variables for easier integration.
- Outcome: Make complex trigonometric integrals more manageable and solvable.
Understanding trigonometric integration and its substitution methods is crucial for breaking down complex trigonometric expressions into easy-to-handle parts.
Definite Integrals
Definite integrals are used to calculate the area under a curve within a specific interval, providing a numerical value. In contrast to indefinite integrals, definite integrals have established limits of integration. However, in this exercise, we tackle definite integrals using substitution to simplify our work, despite not specifying limits exactly in the problem instructions. Yet, understanding the role of definite integrals helps in many practical applications, such as finding real-world areas.
When solving a definite integral, once the substitution simplifies the integral, evaluating it within the given limits is the final step. Even though this exercise primarily handles indefinite forms, remembering definite integrals helps contextualize applications.
- Application: Calculate concrete values for areas beneath curves.
- Comparison to Indefinite: Requires evaluation over an interval, yielding a specific numerical result unlike indefinite integrals.
Indefinite Integrals
Indefinite integrals, unlike their definite counterparts, result in a general form plus a constant \( C \). They symbolize the family of functions that, if differentiated, yield the original integrand. In this exercise, all parts converge to formulas incorporating an indefinite integral. The final answers of the problem involve adding \( C \) after back-substituting transformations, showing that exact original expressions are accounted for.The advantage of indefinite integrals is that they represent general solutions without being tied to specific limits, offering mathematical flexibility:
- Output: A function plus a constant \( C \), representing all possible antiderivatives of an expression.
- Importance: Provides foundational knowledge before exploring specific cases using definite integrals.
- Usage: Often an intermediate step before solving problems involving definite integrals. Allows for evaluating any point within a continuous integral system.
