Question

Use substitution to convert the integrals to integrals of rational functions. Then use partial fractions to evaluate the integrals. \(\int_{1}^{2} \frac{1}{x^{2} \sqrt{4-x^{2}}} d x\)

Short answer

The evaluated integral is \( \frac{\sqrt{3}}{12} \).

Step-by-step solution

Identify a suitable substitution

We notice that the integral involves a square root of the form \( \sqrt{4-x^2} \), which suggests the trigonometric substitution \( x = 2 \sin{\theta} \). This will simplify the square root.

Perform the substitution

Substitute \( x = 2 \sin{\theta} \). Then \( dx = 2 \cos{\theta} \, d\theta \) and \( \sqrt{4-x^2} = \sqrt{4-4\sin^2{\theta}} = 2\cos{\theta} \). The integral bounds change from \( x = 1 \) to \( \theta = \arcsin{\frac{1}{2}} \), and from \( x = 2 \) to \( \theta = \frac{\pi}{2} \).

Simplify the integral using substitution

Substitute into the integral to get: \[ \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{1}{(2\sin{\theta})^2 \cdot 2\cos{\theta}} \cdot 2\cos{\theta} \, d\theta = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{1}{4\sin^2{\theta}} \, d\theta = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{1}{4\sin^2{\theta}} \, d\theta \]. Cancel the factor \( \cos{\theta} \) from the numerator and denominator.

Rewrite the integral in terms of a familiar form

The integral \( \int \frac{1}{\sin^2{\theta}} \, d\theta \) is equivalent to \( \int \csc^2{\theta} \, d\theta \), whose integral is \( -\cot{\theta} \). Thus we rewrite the integral as \[ \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{1}{4} \csc^2{\theta} \, d\theta \].

Integrate and evaluate the bounds

Integrate to get \( -\frac{1}{4}\cot{\theta} \), and now evaluate it from \( \theta = \frac{\pi}{6} \) to \( \theta = \frac{\pi}{2} \): \[ \left[-\frac{1}{4} \cot{\frac{\pi}{2}}\right] - \left[-\frac{1}{4} \cot{\frac{\pi}{6}}\right] = 0 - \left[-\frac{1}{4} \cdot \frac{\sqrt{3}}{3} \right] \].

Simplify and compute the final result

Substitute and simplify to get \( \frac{\sqrt{3}}{12} \). Since \( \cot{\frac{\pi}{2}} = 0 \) and \( \cot{\frac{\pi}{6}} = \frac{\sqrt{3}}{3} \), the expression evaluates to \( \frac{\sqrt{3}}{12} \).

Key concepts

Trigonometric Substitution

In calculus, trigonometric substitution is a powerful technique used to solve integrals involving square roots of quadratic expressions. In our exercise, the integral has a square root of the form \(\sqrt{4-x^2}\). This is a cue for using trigonometric substitution. The main idea is to express \(x\) in terms of a trigonometric function that simplifies the expression under the square root.
For the given integral, we substitute \(x = 2\sin \theta\). This helps transform the square root part \(\sqrt{4-x^2}\) into \(2 \cos \theta\). The corresponding \(dx = 2 \cos \theta \,d \theta\) completes the substitution. This trick effectively reduces the complexity of solving the integral.
This technique is very useful since it changes a difficult integral into one involving trigonometric functions, which are often easier to work with. Remember the key substitutions for these integrals: if you see \(\sqrt{a^2 - x^2}\), think about \( x = a \sin \theta\).

Partial Fractions

Partial fraction decomposition is another important method often used to integrate rational functions. Once you have polynomial expressions in a fraction, you can break them down into simpler fractions. This decomposition allows us to work with easier integrals.
In the context of definite integrals, partial fractions are not directly used in this problem until after trigonometric substitution has simplified the expression. If we had arrived at a rational function, we would apply partial fractions to break it down.
This technique is handy when dealing with integrals of rational functions, which are ratios of polynomials. After decomposing, we integrate each fraction separately. Although not explicitly used in each step of this exercise, partial fractions can simplify integrals and should definitely be on your calculus toolkit.

Definite Integrals

Definite integrals compute the accumulated quantity, like area under a curve, between two limits. In this exercise, we solve \(\int_{1}^{2} \frac{1}{x^{2} \sqrt{4-x^{2}}} \, dx\) with limits. After performing substitution, these are transformed to match the new trigonometric variable.
So when \(x = 1\), \(\theta = \arcsin\left(\frac{1}{2}\right)\), and when \(x = 2\), \(\theta = \frac{\pi}{2}\). These values give us the new bounds for the integral \(\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \).
Substitution aids in transforming the limits and restructuring the integrand for easier integration. After integrating with these bounds, final evaluation tells us the area or the specific value the integral represents between these points. This practice makes definite integrals a critical tool in calculus for real-world applications.

Rational Functions

Rational functions are fractions where both the numerator and the denominator are polynomials. These functions are common in calculus integrals. Often, with complex rational functions, integration techniques like partial fraction decomposition assist in breaking them into simpler parts.
In this problem, though the initial setup wasn't in the clear form of a rational function, after reshaping through trigonometric substitution, it becomes closer to one, because we simplify to forms recognizable such as \(\csc^2 \theta\).
Understanding rational functions is vital, especially when integrating them. Not only does it help in direct computation, but also gives insight into possible techniques for simplification. Recognizing these structures can streamline solving integrals, making calculus problems significantly simpler.