Chapter 7: Problem 22
Use the method of completing the square, along with a trigonometric substitution if needed, to evaluate each integral. \(\int \frac{d x}{\sqrt{16+6 x-x^{2}}}\)
Short Answer
Expert verified
The integral is \(\arcsin\left(\frac{x-3}{5}\right) + C\).
Step by step solution
01
Simplify the Expression
Start by looking at the expression inside the square root which is \(16 + 6x - x^2\). Rewrite it as \(-x^2 + 6x + 16\).
02
Complete the Square
To complete the square for the expression \(-x^2 + 6x + 16\), first factor out a \(-1\): \(-(x^2 - 6x) + 16\). Next, find the value that completes the square inside the parenthesis: take half of \(-6\), which is \(-3\), and square it to get \(9\). Add and subtract \(9\) within the expression: \(-(x^2 - 6x + 9) + 9 + 16\). This simplifies to \(-(x-3)^2 + 25\).
03
Rewrite the Integral Using the Complete Square Form
Substitute the completed square form back into the integral: \(\int \frac{dx}{\sqrt{25 - (x-3)^2}}\).
04
Trigonometric Substitution
Recognize the form \(25 - (x-3)^2\) as \(a^2 - u^2\), where \(a = 5\) and \(u = x-3\). Use the substitution \(u = 5 \sin \theta\), which leads to \(du = 5 \cos \theta \ d\theta\).
05
Substitute and Simplify the Integral
After substitution, the integral becomes \(\int \frac{5 \cos \theta \, d\theta}{\sqrt{25 - 25 \sin^2 \theta}} = \int \frac{5 \cos \theta \, d\theta}{5 \cos \theta}\). This can be simplified to \(\int d\theta\).
06
Evaluate the Integral
The integral of \(d\theta\) is simply \(\theta + C\).
07
Back-Substitute for x
Revert the substitution \(x-3 = 5\sin\theta\) to find \(\theta = \arcsin\left(\frac{x-3}{5}\right)\). Thus, the antiderivative is \(\arcsin\left(\frac{x-3}{5}\right) + C\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Completing the Square
The process of completing the square is a technique used to transform a quadratic expression into a perfect square form. This is particularly helpful in calculus integration because it simplifies complex expressions and aids in making easy substitutions for solving integrals.
To complete the square for an expression like \(-x^2 + 6x + 16\), we begin by factoring out the coefficient of the \(x^2\) term if it isn't 1. In this case, we factor out \(-1\), resulting in \(-(x^2 - 6x)\) + 16\.
Next, identify half of the coefficient of \(-6x\), which is \(-3\). Square this value to get \(9\). Add and subtract this squared number within the expression, giving us the completed square form \(-(x^2 - 6x + 9) + 16 + 9\).
Thus, this expression becomes \(-(x-3)^2 + 25\). The main advantage here is converting the quadratic into a form that reflects a simpler pattern akin to \(a^2 - u^2\), paving the way for trigonometric substitution later.
To complete the square for an expression like \(-x^2 + 6x + 16\), we begin by factoring out the coefficient of the \(x^2\) term if it isn't 1. In this case, we factor out \(-1\), resulting in \(-(x^2 - 6x)\) + 16\.
Next, identify half of the coefficient of \(-6x\), which is \(-3\). Square this value to get \(9\). Add and subtract this squared number within the expression, giving us the completed square form \(-(x^2 - 6x + 9) + 16 + 9\).
Thus, this expression becomes \(-(x-3)^2 + 25\). The main advantage here is converting the quadratic into a form that reflects a simpler pattern akin to \(a^2 - u^2\), paving the way for trigonometric substitution later.
Trigonometric Substitution
Trigonometric substitution is a powerful method used for simplifying integrals especially when dealing with expressions under square roots involving the sum or difference of squares.
After having completed the square, we encountered the expression \(25 - (x-3)^2\). This expression takes the form of \(a^2 - u^2\) where \(a = 5\) and \(u = x-3\).
The goal here is to use a trigonometric identity to simplify the integral. By setting \(u = 5 \sin \theta\), the expression \(25 - 25 \sin^2 \theta\) turns into \(25 \cos^2 \theta\) given the identity \(1 - \sin^2 \theta = \cos^2 \theta\).
This substitution simplifies the squareroot to \(5 \cos \theta\), making the integral much easier to solve, by reducing it to a basic integral in terms of \theta\. This step highlights how trigonometric substitutions condense seemingly complex integrals into manageable forms.
After having completed the square, we encountered the expression \(25 - (x-3)^2\). This expression takes the form of \(a^2 - u^2\) where \(a = 5\) and \(u = x-3\).
The goal here is to use a trigonometric identity to simplify the integral. By setting \(u = 5 \sin \theta\), the expression \(25 - 25 \sin^2 \theta\) turns into \(25 \cos^2 \theta\) given the identity \(1 - \sin^2 \theta = \cos^2 \theta\).
This substitution simplifies the squareroot to \(5 \cos \theta\), making the integral much easier to solve, by reducing it to a basic integral in terms of \theta\. This step highlights how trigonometric substitutions condense seemingly complex integrals into manageable forms.
Definite Integral
The concept of a definite integral extends beyond just finding the area under a curve; it's about evaluating an antiderivative over specific limits. However, the current exercise focuses solely on the indefinite integral, meaning the solution describes a general antiderivative form.
Once the expression is transformed through trigonometric substitution, the definite integration process involves evaluating the resulting function over a range. But in indefinite terms, which is the focus here, the solution aims to resolve into an antiderivative.
In this exercise, the integration process simplifies significantly upon reducing the integral to \int d\theta\, the result of which is \theta + C\ followed by back-substitution to express \theta\ in terms of the original variable \(x\).
The step-by-step back-substitution, \(x-3 = 5 \sin \theta\), gives us the inverse operation \( heta = \arcsin\(\frac{x-3}{5}\)\), thus expressing the indefinite integral in terms of the original variable. It's crucial to understand this overall strategy for translating complex integrals into forms suitable for calculation before finally returning to familiar variable settings.
Once the expression is transformed through trigonometric substitution, the definite integration process involves evaluating the resulting function over a range. But in indefinite terms, which is the focus here, the solution aims to resolve into an antiderivative.
In this exercise, the integration process simplifies significantly upon reducing the integral to \int d\theta\, the result of which is \theta + C\ followed by back-substitution to express \theta\ in terms of the original variable \(x\).
The step-by-step back-substitution, \(x-3 = 5 \sin \theta\), gives us the inverse operation \( heta = \arcsin\(\frac{x-3}{5}\)\), thus expressing the indefinite integral in terms of the original variable. It's crucial to understand this overall strategy for translating complex integrals into forms suitable for calculation before finally returning to familiar variable settings.
