Problem 21 - Calculus of a Single Variable

Question

Evaluating a Definite Integral In Exercises \(21-32\) evaluate the definite integral. $$ \int_{0}^{1 / 6} \frac{3}{\sqrt{1-9 x^{2}}} d x $$

Step-by-Step Solution

Answer

The final answer after evaluating the definite integral is \(\frac{1}{2}\)

Step 1: Perform Trigonometric Substitution

Let's substitute \(x = \frac{1}{3}\sin(\theta)\). This implies that \(dx = \frac{1}{3}\cos(\theta)d\theta\). When \(x = 0\), \(\theta = 0\) and when \(x = 1/6\), \(\theta = \pi/6\). The given integral then transforms to \(\int_{0}^{\pi/6} \frac{3}{\sqrt{1-9(\frac{1}{3}\sin(\theta))^2}}*\frac{1}{3}\cos(\theta)d\theta\)

Step 2: Simplify the Integral

After substituting, simplifying the equations gives us the new integral as \(\int_{0}^{\pi/6} \cos(\theta) d\theta\)

Step 3: Evaluate the Integral

We can evaluate the integral using the antiderivative of \(cos(\theta)\), which is \(\sin(\theta)\). Thus the evaluation becomes \(\sin(\theta)\Big|_0^{\pi/6}\)

Step 4: Substitute the Limits

Inserting the upper and lower limits, we get \(\sin(\pi/6) - \sin(0)\)

Key Concepts

Trigonometric Substitution

When faced with challenging integrals that involve roots, such as \( \sqrt{a^2 - x^2} \), \( \sqrt{a^2 + x^2} \), or \( \sqrt{x^2 - a^2} \), trigonometric substitution can be a powerful technique. This method leverages the Pythagorean identities within trigonometry to simplify these roots.

In our exercise, the presence of \( \sqrt{1 - 9x^2} \) suggests that a trigonometric substitution might be helpful. We note that \( \sin^2(\theta) + \cos^2(\theta) = 1 \) can be rearranged to \( \cos^2(\theta) = 1 - \sin^2(\theta) \), which looks quite similar to the expression under the root in our integral.

The substitution \( x = \frac{1}{3}\sin(\theta) \) is chosen to match the structure of the root expression, transforming the integrand into a trigonometric function. This substitution not only simplifies the root but also changes the differential \( dx \) to a trigonometric differential \( \frac{1}{3}\cos(\theta)d\theta \)—a form that is often easier to integrate.

Integral Simplification

After applying a trigonometric substitution, the integral often still needs to be simplified further to make it evaluable. This process involves exploiting trigonometric identities or algebraic manipulation.

In our example, the simplified integral \( \int_{0}^{\pi/6} \cos(\theta) d\theta \) emerges from the substitution. The simplification comes from recognizing that the trigonometric identity used in the substitution cancels out the square in \( 1 - 9x^2 \) and that \( \frac{3}{1} \times \frac{1}{3} = 1 \), leaving us with the cos function.

This simplification reduces the original integral, which contained a root and a function squared, to a much simpler form involving only a trigonometric function—cosine in this case. It is this simpler form that we can readily evaluate using our knowledge of trigonometric antiderivatives.

Antiderivatives

Antiderivatives, also known as indefinite integrals, are the reverse of derivatives. They are used to find a function whose derivative is the given function. This process play a crucial role in evaluating definite integrals, where we seek the net area under a curve.

In our exercise, once the integral has been simplified to \( \int_{0}^{\pi/6} \cos(\theta) d\theta \) through substitution and simplification, we need to find the antiderivative of \( \cos(\theta) \). The antiderivative of cosine is \( \sin(\theta) \), because \( \frac{d}{d\theta} \sin(\theta) = \cos(\theta) \).

After finding the antiderivative, the next step is to apply the Fundamental Theorem of Calculus, which states that if F is an antiderivative of f on an interval I, then for any a and b in I, \(\int_a^b f(x) dx = F(b) - F(a)\). By substituting \( \pi/6 \) and \( 0 \) into the antiderivative, we obtain the value of the definite integral, which represents the area under the cosine curve from \( \theta = 0 \) to \( \theta = \pi/6 \) in the context of the problem.