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Magazine Chapter 9: Problem 6
Evaluate the integral. $$ \int \frac{1}{\sqrt{7+6 x-x^{2}}} d x $$
Short Answer
Expert verified
The evaluated integral is \( \sin^{-1}\left( \frac{x-3}{4} \right) + C \).
Step by step solution
01
Identify the Integral as an Inverse Trigonometric Substitution
Recognize that the integral \( \int \frac{1}{\sqrt{7+6x-x^2}} \, dx \) resembles the form \( \int \frac{1}{\sqrt{a^2-x^2}} \, dx \), which suggests using a trigonometric substitution for inverse sine. First, complete the square in the expression \( 7+6x-x^2 \).
02
Complete the Square
Rewrite \( 7 + 6x - x^2 \) as a perfect square by completing the square. Start by rewriting it as \( -(x^2 - 6x - 7) \). Add and subtract \( 9 \) (i.e., \( (6/2)^2 \)) to complete the square:\[-(x^2 - 6x - 9 + 9 - 7) = -( (x-3)^2 - 16) = 16 - (x-3)^2.\]
03
Rewrite the Integral Using the Complete Square
Using the completed square from the previous step, rewrite the integral:\[ \int \frac{1}{\sqrt{16 - (x-3)^2}} \, dx.\]This can now be integrated using the substitution \( x - 3 = 4\sin\theta \).
04
Perform the Trigonometric Substitution
Set \( x - 3 = 4\sin\theta \). Then, \( dx = 4\cos\theta \, d\theta \). Substitute these into the integral:\[ \int \frac{1}{\sqrt{16 - 16\sin^2\theta}} \cdot 4\cos\theta \, d\theta = \int \frac{4\cos\theta}{4\cos\theta} \, d\theta = \int d\theta.\]
05
Integrate and Back-Substitute
Integrate \( \int d\theta \) to get \( \theta + C \). Since \( x - 3 = 4\sin\theta \), use \( \theta = \sin^{-1}\left( \frac{x-3}{4} \right) \), yielding:\[\theta = \sin^{-1}\left( \frac{x-3}{4} \right) + C.\]
06
State the Final Answer
The evaluated integral is:\[ \int \frac{1}{\sqrt{7+6x-x^2}} \, dx = \sin^{-1}\left( \frac{x-3}{4} \right) + C.\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Inverse Trigonometric Substitution
Inverse trigonometric substitution is a powerful technique used in calculus to evaluate certain types of integrals.
When faced with an integral that resembles the form \( \int \frac{1}{\sqrt{a^2-x^2}} \, dx \), we can use this method to simplify the problem.
By recognizing the structure, we can substitute the variable with a trigonometric function that simplifies the expression inside the square root.
When faced with an integral that resembles the form \( \int \frac{1}{\sqrt{a^2-x^2}} \, dx \), we can use this method to simplify the problem.
By recognizing the structure, we can substitute the variable with a trigonometric function that simplifies the expression inside the square root.
- It is particularly helpful when dealing with expressions that involve square roots of quadratic polynomials.
- The goal is to transform the integral into one that is easier to evaluate, typically resulting in an inverse trigonometric function like \( \sin^{-1} \) or \( \cos^{-1} \).
Completing the Square
Completing the square is a technique used to simplify quadratic expressions, making them easier to handle in calculus.
When you have a quadratic like \( 7 + 6x - x^2 \), it can often be rewritten in a form that resembles a squared term, plus or minus a constant.
When you have a quadratic like \( 7 + 6x - x^2 \), it can often be rewritten in a form that resembles a squared term, plus or minus a constant.
- The process involves rearranging and factoring the quadratic to find a perfect square trinomial.
- This is done by taking half of the linear coefficient \( b \, ( \frac{6}{2} = 3) \), squaring it \( 9 \), then adding and subtracting it within the expression.
- In this problem, the completed square becomes \( 16 - (x - 3)^2 \).
Definite Integrals
Definite integrals are important in calculus as tools for calculating areas under curves.
The integral \( \int \, dx \), when given limits, yields a specific numerical value.
The integral \( \int \, dx \), when given limits, yields a specific numerical value.
- Unlike indefinite integrals, which result in a family of functions, definite integrals give you a precise area.
- They're often used to find areas, volumes, and other quantities that accumulate over intervals.
Trigonometric Substitution
Trigonometric substitution is a technique that involves replacing a variable with a trigonometric expression.
This technique simplifies integrals involving the square roots of quadratic expressions.
This technique simplifies integrals involving the square roots of quadratic expressions.
- After completing the square, one common substitution is \( x - a = b\sin\theta \), where \( b \) is the square root of the constant term.
- This alters the integral into a simpler form, often eliminating the square root through trigonometric identities.
