Question

$$\text {Evaluate the following integrals.}$$ $$\int_{-\pi / 3}^{\pi / 3} \sqrt{\sec ^{2} \theta-1} d \theta$$

Short answer

Question: Evaluate the following integral: $$\int_{-\pi / 3}^{\pi / 3} \sqrt{\sec ^{2} \theta-1} d \theta$$ Answer: The value of the integral is $$\frac{3}{2}.$$

Step-by-step solution

Simplify the integrand

We know that: $$\sec^2\theta - 1 = \tan^2\theta$$ So, the given integral can be rewritten as: $$\int_{-\pi/3}^{\pi/3} \sqrt{\tan^2\theta} d\theta$$ Since \(\tan \theta\) is an odd function, we have: $$\int_{-\pi/3}^{\pi/3} \sqrt{\tan^2\theta} d\theta = \int_{-\pi/3}^{\pi/3} |\tan\theta| d\theta$$

Apply the substitution method

Let's use the substitution method with: $$u = \tan\theta \Rightarrow du = \sec^2\theta d\theta$$ In terms of \(u\), the limits of integration change as well. When \(\theta=-\pi/3\), \(u=-\sqrt{3}\), and when \(\theta=\pi/3\), \(u=\sqrt{3}\). Hence, our integral becomes: $$\int_{-\sqrt{3}}^{\sqrt{3}} |u| du$$ Now we can split the integral into two parts: $$\int_{-\sqrt{3}}^{\sqrt{3}} |u| du = \int_{-\sqrt{3}}^{0}-u du + \int_{0}^{\sqrt{3}}u du$$ Calculating these integrals, we have: $$\begin{aligned} \int_{-\sqrt{3}}^{0}-u du + \int_{0}^{\sqrt{3}}u du &= -\frac{1}{2} u^2\Big|_{-\sqrt{3}}^{0} + \frac{1}{2}u^2\Big|_{0}^{\sqrt{3}} \\ &= -\frac{1}{2}(0^2-(-\sqrt{3})^2) + \frac{1}{2}(\sqrt{3}^2 - 0^2)\\ &= -\frac{1}{2}(0 - 3) + \frac{1}{2}(3 - 0)\\ &= \frac{3}{2} \end{aligned} $$ So the final answer is: $$\int_{-\pi / 3}^{\pi / 3} \sqrt{\sec ^{2} \theta-1} d \theta = \frac{3}{2}$$

Key concepts

Trigonometric Substitution

Trigonometric substitution is a powerful method used to evaluate integrals involving radicals or algebraic expressions. It simplifies the integral by substituting a trigonometric function to replace a variable, transforming the integral into a trigonometric one which is often easier to solve.
In the given problem, we start by noticing we have \( \sec^2\theta - 1 \),which can be transformed using trigonometric identities. Since \( \sec^2\theta = 1 + \tan^2 \theta \), the expression simplifies to\( \tan^2 \theta \).
This substitution not only simplifies the original integrand but links it to the well-known tangent trig function, clearing the path for further simplifications or transformations needed for evaluating the integral.

Odd Functions

Odd functions have a special property related to definite integrals. An odd function satisfies the condition \( f(-x) = -f(x) \).When an odd function is integrated over a symmetric interval like \([-a, a]\),its integral equals zero, provided the function does not have an absolute value marker.
In the exercise, \( \tan \theta \)is identified as an odd function. Consequently, \( \sqrt{\tan^2\theta} \)translates to\( |\tan\theta| \) due to the absolute value. Without the absolute values, the integral over \([-\pi/3, \pi/3]\) would have been zero, leveraging symmetry. However, the presence of the absolute value inverts negative results to positive, requiring partitioning into two separate integrals for evaluation.

Absolute Value

The absolute value function, denoted as \( |x| \), takes any real number and returns its non-negative counterpart. This is particularly important when dealing with symmetry over integrals of odd functions. In the integral \( \int_{-\pi/3}^{\pi/3} |\tan \theta| d\theta \),the absolute value ensures that all areas under the curve are accumulated as positive, regardless of their initial negative nature.
When the trigonometric substitution involved \( \tan \theta \), the integrand became \( |u| \) after substitution. This prompted the separation of the integral at zero into intervals:

• One from \(-\sqrt{3}\)to zero,
• and another from zero to \(\sqrt{3}\).

The absolute value accounts for the actual magnitude swept by the curve, orienting all parts of the area into a positive integration result.

Algebraic Simplification

Algebraic simplification involves breaking down expressions into simpler or more calculable forms, which is crucial for ease in integration. The expression \( \sec^2\theta - 1 = \tan^2\theta \) is a direct application of trigonometric identities, aligning expressions into recognizable forms.
Within this exercise, simplification was a key in transitioning from a complex radical to an absolute function of \( \tan \theta \).Once the integral transitioned into one concerning \( |u| \),the expression was further divided and expressed by straightforward degree polynomial integrals like \(-u\) and \(u \).
This bifurcation using simple power rules allowed us to calculate each segment effectively and quickly sum them to arrive at the final answer. Attaining \(3/2\) demonstrates the essential role of simplification in solving integrals efficiently.