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Magazine Chapter 8: Problem 11
Trigonometric substitutions Evaluate the following integrals using trigonometric substitution. $$\int_{1 / 2}^{\sqrt{3} / 2} \frac{x^{2}}{\sqrt{1-x^{2}}} d x$$
Short Answer
Expert verified
Based on the step by step solution provided, answer the following:
**Question:** Determine the value of the following integral:
$$\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \frac{x^2}{\sqrt{1-x^2}} dx$$
**Answer:** The value of the integral is:
$$\frac{1}{4}\left(\pi + \frac{\sqrt{3}}{2}\right)$$
Step by step solution
01
Determine the appropriate trigonometric substitution
Since we have a \(\sqrt{1-x^2}\) term in the integral, we can rewrite the expression using the sine function. We can do this by introducing a substitution \(x = \sin{\theta}\) for some angle \(\theta\). Thus, we have:
$$
x = \sin{\theta} \Rightarrow x^2 = \sin^2{\theta} \Rightarrow 1-x^2 = 1-\sin^2{\theta} = \cos^2{\theta}.
$$
02
Find the differential
Now we need the differential \(dx\) in terms of \(d\theta\). Since \(x = \sin{\theta}\), we have:
$$dx = \frac{d(\sin\theta)}{d\theta}d\theta = \cos{\theta} d\theta.$$
03
Substitute the expressions and change the integration bounds
We substitute \(x=\sin{\theta}\) and \(dx=\cos{\theta} d\theta\) into our original integral. We also change the integration bounds to correspond to the new variable. When \(x=\frac{1}{2}\), we have \(\sin{\theta} = \frac{1}{2}\), so \(\theta = \frac{\pi}{6}\). When \(x=\frac{\sqrt{3}}{2}\), we have \(\sin{\theta} = \frac{\sqrt{3}}{2}\), so \(\theta = \frac{\pi}{3}\). Our integral now becomes:
$$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin^2{\theta}}{\sqrt{\cos^2{\theta}}} \cos{\theta} d\theta$$
04
Simplify the integral
Inside the integral, we have \(\frac{\sin^2{\theta} \cos{\theta}}{\sqrt{\cos^2{\theta}}}\). The \(\cos^2{\theta}\) inside the square root will simplify to \(|\cos{\theta}|\). Since we are in the first quadrant (\(\frac{\pi}{6} \leq \theta \leq \frac{\pi}{3}\)), the cosine is positive. Therefore, we can simplify the expression:
$$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sin^2{\theta} d\theta$$
05
Use the double-angle identity
To make this integral easier to evaluate, we can use the double-angle identity for \(\sin^2{\theta}\):
$$\sin^2{\theta}=\frac{1-\cos(2\theta)}{2}.$$
Substitute this into our integral:
$$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1-\cos(2\theta)}{2} d\theta$$
06
Evaluate the integral
Since our integral now consists of two simple terms, we can easily evaluate it. First, integrate with respect to \(\theta\):
$$I = \frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} (1-\cos(2\theta)) d\theta = \frac{1}{2} \left(\theta -\frac{1}{2}\sin(2\theta) \right) \Bigg|_{\frac{\pi}{6}}^{\frac{\pi}{3}}$$
07
Calculate the result
Finally, plug in the integration bounds into the expression:
$$I = \frac{1}{2}\left(\left(\frac{\pi}{3}-\frac{1}{2}\sin\left(\frac{2\pi}{3}\right)\right) - \left(\frac{\pi}{6}-\frac{1}{2}\sin\left(\frac{\pi}{3}\right)\right)\right)$$
Calculating the sine values and simplifying, we find the final result:
$$I = \frac{1}{4}\left(\pi + \frac{\sqrt{3}}{2}\right)$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integral
In calculus, the definite integral of a function gives us the area under the graph of the function, bounded between two specified points. The process of finding this integral is known as integration. Specifically, for a function f(x), the definite integral from a to b is denoted as \[\int_a^b f(x) dx.\] It's a fundamental concept as it touches on quantities like area, volume, and the accumulated sum of continuously changing rates.
When we look at an integral like \[\int_{1 / 2}^{\sqrt{3} / 2} \frac{x^{2}}{\sqrt{1-x^{2}}} dx,\] we are essentially seeking the total sum of infinitely small pieces of the function's graph between the two bounds 1/2 and \sqrt{3}/2. This kind of problem often requires advanced methods to solve, such as trigonometric substitution, which tactically transforms the integral into a more manageable form.
When we look at an integral like \[\int_{1 / 2}^{\sqrt{3} / 2} \frac{x^{2}}{\sqrt{1-x^{2}}} dx,\] we are essentially seeking the total sum of infinitely small pieces of the function's graph between the two bounds 1/2 and \sqrt{3}/2. This kind of problem often requires advanced methods to solve, such as trigonometric substitution, which tactically transforms the integral into a more manageable form.
Integration Techniques
There are various integration techniques used to evaluate complex integrals. Techniques such as substitution, integration by parts, partial fractions, and trigonometric substitution are indispensable tools in a mathematician's toolkit.
For the integral given in our example, trigonometric substitution is ideal because of the square root of 1 - x虏. This can be related to the Pythagorean identity 蝉颈苍虏(胃) + cos虏(胃) = 1, allowing us to rewrite the integral in terms of trigonometric functions.
To perform a trigonometric substitution,
For the integral given in our example, trigonometric substitution is ideal because of the square root of 1 - x虏. This can be related to the Pythagorean identity 蝉颈苍虏(胃) + cos虏(胃) = 1, allowing us to rewrite the integral in terms of trigonometric functions.
To perform a trigonometric substitution,
Identify the Substitution
First, select the correct trigonometric function to substitute based on the form of the integrand. In our case, it's the sine function. After the substitution, find the corresponding differential. In this example, we obtained dx = cos(胃)d胃 after substituting x = sin(胃).Next,Change the Limits
Convert the integral's limits to match the new variable. Here, we replaced x limits with 胃 limits. With that done, we acquire a new integrand in terms of trigonometric functions, which often simplifies the integration process.Double-Angle Identities
The double-angle identities are specific cases of the sum formulas in trigonometry and are very useful in simplifying expressions involving trigonometric functions. The double-angle identities express sin(2x), cos(2x), and tan(2x) in terms of sin(x), cos(x), and tan(x).
The double-angle formula for the sine function is \[\sin(2x) = 2\sin(x)\cos(x),\] and can be used in reverse to express 蝉颈苍虏(虫) as \[\sin^2(x) = \frac{1 - \cos(2x)}{2}.\] This expression is highlighted in the step-by-step solution, where it simplifies 蝉颈苍虏(胃) into a form that is easier to integrate. Using this identity ultimately enables the evaluation of the definite integral without the need to wrestle with more complex trigonometric integrals directly.
Mastering the use of double-angle identities in calculus is highly beneficial, as they frequently allow for the transformation of difficult integrals into more standard forms that are simpler to integrate, especially when combined with other methods like trigonometric substitution.
The double-angle formula for the sine function is \[\sin(2x) = 2\sin(x)\cos(x),\] and can be used in reverse to express 蝉颈苍虏(虫) as \[\sin^2(x) = \frac{1 - \cos(2x)}{2}.\] This expression is highlighted in the step-by-step solution, where it simplifies 蝉颈苍虏(胃) into a form that is easier to integrate. Using this identity ultimately enables the evaluation of the definite integral without the need to wrestle with more complex trigonometric integrals directly.
Mastering the use of double-angle identities in calculus is highly beneficial, as they frequently allow for the transformation of difficult integrals into more standard forms that are simpler to integrate, especially when combined with other methods like trigonometric substitution.
