Problem 17 - Essential Calculus

Question

Evaluate the integral. $$ \int_{0}^{1 / 6} \frac{1}{\sqrt{1-9 x^{2}}} d x $$

Step-by-Step Solution

Answer

The value of the integral is \(\pi/18\).

Step 1: Identify the substitution

Note that the denominator of the integral function has the form of a ^2 - b ^2, which suggests the use of the substitution method for integration with \(u = 3x\). The differential of u \(du\) is 3 times the differential of x, dx, so \(dx = du/3\).

Step 2: Substitute and Simplify

Substituting \(u = 3x\) and \(dx = du/3\) into the integral gives us \(\int_{0}^{1 / 2} \frac{1}{3\sqrt{1-u^{2}}} du\). The 1/3 can be taken out of the integral leading to \(1/3 \int_{0}^{1 / 2} \frac{1}{\sqrt{1-u^{2}}} du\). The integral is now in a form which is the standard integral \(\int \frac{1}{\sqrt{1-u^{2}}} du = arcsin(u) + C\).

Step 3: Solve the Integral

Using the formula, the integral translates into \(1/3 arcsin(u)\) from 0 to 1/2.

Step 4: Calculate the Definite Integral

Definite integral means we subtract the function evaluated at the lower limit from the function evaluated at the upper limit, i.e. \(1/3[\arcsin(1/2) - \arcsin(0)] = 1/3 * [\pi/6 - 0]= \pi/18\)

Key Concepts

Substitution Method

The substitution method is a powerful technique used to simplify the process of finding definite integrals, especially when the integral involves a composite function. Imagine it as a "u-turn," where you replace a complex part of the function with a simpler variable like "u." This makes the integration easier to handle.
The first step in using the substitution method is identifying a suitable substitution. This usually involves recognizing a formula or pattern within the integral. In our example, you see a square root of the form \(1 - 9x^2\).
To make the function simpler, you substitute \(u = 3x\). This choice is strategic, often because substituting clears up the complex parts of the integral. Here, differentiating gives us \(du = 3 \, dx\), or rearranging: \(dx = \frac{du}{3}\).
Once substitution is done, the integral transforms, making the expression more straightforward to integrate. That's the magic of a good substitution! It alters the integral into a well-known form that's easier to evaluate.

Definite Integral Evaluation

Definite integrals provide the exact area under a curve between two specific points on the x-axis, providing a precise number rather than a function. Let's break down the definite integral evaluation steps.
After performing the substitution, the integral simplified to a familiar form: \(1/3 \int_{0}^{1/2} \frac{1}{\sqrt{1-u^{2}}} \, du\). The substitution changed the limits of integration too, from \(x = 0\) and \(x = 1/6\) to \(u = 0\) and \(u = 1/2\).
Next, the integral \(\int \frac{1}{\sqrt{1-u^2}} \, du\) is recognized as the derivative of the inverse sine function, or \(\arcsin(u)\). Thus, this transforms our integral into \(1/3 [\arcsin(u)]\) evaluated from 0 to 1/2.
With definite integrals, the real work is completed with substitution back and plugging in the boundaries. Thus, we calculate the difference: \(1/3 [\arcsin(1/2) - \arcsin(0)]\). The evaluated result is \(\pi/18\), which gives the precise area under the curve between the original limits.

Trigonometric Integrals

Trigonometric integrals include integrals that involve trigonometric functions. They often appear more complex than they truly are.
In our example, after a substitution, the integral simplifies to a form involving \(\arcsin\), specifically \(\int \frac{1}{\sqrt{1-u^{2}}} \, du\). Understanding this integral as a key trigonometric derivative is crucial.
The integral \(\int \frac{1}{\sqrt{1-u^2}} \, du\) corresponds to \(\arcsin(u) + C\). It's a standard result in calculus, born from the derivative of \(\sin^{-1}(u)\) being \(\frac{1}{\sqrt{1-u^2}}\).
Recognizing these patterns helps in solving trigonometric integrals efficiently. Mastery over these standard forms enables quick identification and solution of related problems, turning cumbersome integrals into manageable tasks.