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Magazine Chapter 26: Problem 742
Integrate the expression: \(\int\left[(\mathrm{x} \mathrm{dx}) /\left\\{\sqrt{\left. \left.\left(\mathrm{ax}-\mathrm{x}^{2}\right)\right\\}\right]}, \quad \mathrm{a}>0\right.\right.\)
Short Answer
Expert verified
The short answer to the integration is: \[-a\sqrt{1-\left(\frac{x}{a}\right)^2} + C\]
Step by step solution
01
Formulate the integral
First, let us write down the integral that we want to calculate: \[\int\frac{x\,dx}{\sqrt{ax-x^2}},\quad a>0\]
02
Perform substitution
Let us perform a substitution using a sine function: Let \(x = a\sin(\theta)\), with \(0 \leq \theta \leq \pi\). This choice will allow us to simplify the expression inside the square root. Now, compute the differential related to x: \[dx = a\cos(\theta) d\theta\]
Now, we can rewrite the integral as: \[\int\frac{a\sin(\theta)(a\cos(\theta) d\theta)}{\sqrt{a(1-\sin^2(\theta))}}\]
03
Simplify the integral
First, notice that the square root denominator can be further simplified: \[\sqrt{a(1-\sin^2(\theta))} = \sqrt{a\cos^2(\theta)} = \cos(\theta)\sqrt{a}\]
Now, we can rewrite our integral as: \[\int\frac{a^2\sin(\theta)\cos(\theta) d\theta}{\cos(\theta)\sqrt{a}}\]
Now, we see that one of the cosine terms in the numerator and denominator cancels out. Then, we are left with: \[\int a\sin(\theta) d\theta\]
04
Integration
Now, we can easily integrate the resulting expression with respect to \(\theta\), obtaining: \[a(-\cos(\theta)) + C\]
05
Reverse substitution
We have obtained the result in terms of \(\theta\), now let's rewrite it in terms of x using the reverse substitution: \[-a\cos(\theta) + C = -a\cos\left(\arcsin\left(\frac{x}{a}\right)\right) + C\]
Recall the trigonometric identity: \[\cos(\arcsin x) = \sqrt{1-x^2}\]
Thus, the result of the integration is: \[-a\sqrt{1-\left(\frac{x}{a}\right)^2} + C\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Trigonometric Substitution
Trigonometric substitution is a technique used in integration when dealing with expressions involving square roots of the form \(\sqrt{a^2 - x^2}\), \(\sqrt{x^2 - a^2}\), or \(\sqrt{x^2 + a^2}\).
This method is particularly useful as it simplifies the integration process by turning complex algebraic expressions into trigonometric ones. To effectively use trigonometric substitution:
This method is particularly useful as it simplifies the integration process by turning complex algebraic expressions into trigonometric ones. To effectively use trigonometric substitution:
- Identify the form of the expression under the square root, which will guide you to choose the correct trigonometric substitution.
- For expressions like \(\sqrt{a^2 - x^2}\), use \(x = a\sin\theta\) because \(\sin^2\theta + \cos^2\theta = 1\) holds true.
- For \(\sqrt{x^2 - a^2}\), try \(x = a\sec\theta\) as \(\sec^2\theta - 1 = \tan^2\theta\).
- For \(\sqrt{x^2 + a^2}\), use \(x = a\tan\theta\) as \(\tan^2\theta + 1 = \sec^2\theta\).
Definite Integral
A definite integral helps calculate the area under a curve within specified bounds. When performed, it eliminates the variable of integration and gives a constant that represents the signed area.
Here鈥檚 what you need to know about definite integrals:
Here鈥檚 what you need to know about definite integrals:
- Defining bounds \(a\) to \(b\) on the integral symbol tells us that we are evaluating the integral from \(x = a\) to \(x = b\).
- They provide a numerical value, in contrast to indefinite integrals which yield an expression plus an arbitrary constant \(C\).
- In the case where a function is above the x-axis, the definite integral gives the area beneath it. If below, the definite integral gives the negative of that area.
Integration by Parts
Integration by parts is a powerful technique derived from the product rule for differentiation. It is useful for finding integrals of products of functions. The formula for integration by parts is:\[\int u \, dv = uv - \int v \, du\]This method involves choosing parts of the integrand to form \(u\) and \(dv\). Here鈥檚 how to use integration by parts effectively:
- If one part of the integrand simplifies upon differentiation and another doesn't complicate upon integration, let the first be \(u\) and the latter \(dv\).
- Differentiate \(u\) to get \(du\) and integrate \(dv\) to get \(v\).
- Substitute back into the integration by parts formula.
Calculus Problem Solving
Calculus problem solving is an approach that combines calculus concepts to solve complex mathematical problems. This usually involves a strategic combination of different techniques and a deep understanding of both integrals and derivatives.
To be proficient in calculus problem solving:
To be proficient in calculus problem solving:
- Understand the behavior of functions through derivatives and integrals. This includes slope determination and area calculation under curves.
- Practice combining techniques such as substitution, integration by parts, and partial fractions. Each method works best with specific forms of expressions.
- Simplify integrands before integration, whenever possible. This might involve algebraic manipulation, trigonometric identities, or completing the square.
- Use graphical interpretations to cross-verify solutions and gain insight into the behavior of the function in question.
