Direct Comparison Test: Integrating Tan(4θ) From 0 To Π/8

Oct 20, 2025 by ADMIN 58 views

Hey guys! Today, we're diving deep into the fascinating world of calculus, specifically focusing on integration using the Direct Comparison Test. We'll be tackling a concrete example: evaluating the integral of tan(4θ) from 0 to π/8. This is super exciting because it combines trigonometric functions, definite integrals, and a powerful convergence test. So, buckle up and let's get started!

Understanding the Direct Comparison Test

Before we jump into the integral itself, let's quickly recap the Direct Comparison Test. This test is a fantastic tool for determining whether an improper integral converges or diverges. Basically, it works by comparing our given integral with another integral whose convergence or divergence is already known.

The key idea is this: If our integral's function is smaller than a convergent integral's function, then our integral also converges. Conversely, if our integral's function is larger than a divergent integral's function, then our integral also diverges. Think of it like comparing the sizes of pizzas – if a smaller pizza is enough to feed everyone, then a much larger pizza is sure to be as well, right? But let's formalize it a little more for the math folks in the room.

Here's the formal statement: Suppose we have two functions, f(x) and g(x), that are continuous and non-negative on the interval [a, ∞). Let's say that 0 ≤ f(x) ≤ g(x) for all x ≥ a. Then:

If ∫[a, ∞) g(x) dx converges, then ∫[a, ∞) f(x) dx also converges.
If ∫[a, ∞) f(x) dx diverges, then ∫[a, ∞) g(x) dx also diverges.

This test relies heavily on choosing the right comparison function, which can sometimes be a bit tricky but that's also what makes it fun. Choosing the right comparison function is crucial. It's like finding the perfect ingredient for a recipe – get it right, and everything comes together beautifully. Get it wrong, and well, you might end up with a mathematical mess. We need a function that is both comparable to our original function and whose integral's convergence or divergence we already know. Common comparison functions often involve powers of x or exponentials, but in our case, since we're dealing with a trigonometric function, we need to think a little differently. The secret sauce here is understanding the behavior of trigonometric functions within the given interval and leveraging their properties to find a suitable comparison. Practice makes perfect in choosing these functions, so don't worry if it doesn't click immediately. Keep exploring different examples, and you'll soon develop a knack for it!

Setting Up Our Integral: ∫[0, π/8] tan(4θ) dθ

Now, let's focus on our specific integral: ∫[0, π/8] tan(4θ) dθ. The first thing we need to do is identify any potential issues. Remember, the Direct Comparison Test applies to improper integrals, which are integrals that either have infinite limits of integration or have discontinuities within the interval of integration.

In our case, we need to check if tan(4θ) has any discontinuities on the interval [0, π/8]. Recall that tan(x) = sin(x) / cos(x), so it's discontinuous wherever cos(x) = 0. Translating this to our function, tan(4θ) is discontinuous when cos(4θ) = 0. Let's find out where this happens within our interval.

Cos(4θ) = 0 when 4θ = π/2 + nπ, where n is an integer. Solving for θ, we get θ = π/8 + nπ/4. Notice that when n = 0, we have θ = π/8, which is the upper limit of our integration! This means we have a discontinuity at θ = π/8, making this an improper integral. You see, identifying these points of discontinuity is like performing a health check on our integral. It helps us understand its behavior and ensures we apply the right techniques for solving it. Overlooking a discontinuity can lead to incorrect results, so it’s a crucial step in the process. In our case, the discontinuity at π/8 signals that we need to approach this integral with a little more care and potentially use limits to evaluate it properly.

Finding a Suitable Comparison Function

This is where the magic happens! We need to find a function that we can compare to tan(4θ) on the interval [0, π/8]. A good starting point is to think about the behavior of tan(x) near x = 0. We know that for small x, tan(x) ≈ x. So, we can expect tan(4θ) to behave similarly to 4θ near θ = 0.

However, we need to be careful. The approximation tan(x) ≈ x only holds for small x. As x gets closer to π/2, tan(x) approaches infinity. We need to find a function that either bounds tan(4θ) from above or below, and whose integral we can easily evaluate. For θ values within the interval [0, π/8], the tangent function starts at 0 and increases. Let's analyze the behavior near our point of discontinuity, π/8. As θ approaches π/8, tan(4θ) approaches infinity. This suggests that we might need to find a function that also diverges to properly compare.

So, let's consider the inequality tan(x) ≥ x for x in [0, π/4). This is a crucial insight because it provides us with a lower bound for the tangent function within a specific interval. Now, applying this inequality to our case, we have tan(4θ) ≥ 4θ for 4θ in [0, π/4), which means θ is in [0, π/16). Unfortunately, this doesn’t cover our entire interval [0, π/8]. We need a comparison that holds over the entire interval.

What about comparing tan(4θ) to a constant multiple of 1/(π/8 - θ)? As θ approaches π/8, this function also approaches infinity. It seems promising, but we need to rigorously show that tan(4θ) is greater than or less than our comparison function. This is where careful analysis and sometimes a bit of trial and error come into play. The goal is to find a comparison function that not only mimics the behavior of the original function near the point of discontinuity but also allows us to easily evaluate the integral.

Applying the Limit Comparison Test (Alternative Approach)

Since finding a suitable function for the Direct Comparison Test can be tricky, let's consider an alternative: the Limit Comparison Test. This test is particularly useful when we're dealing with functions that are hard to directly compare, but their asymptotic behavior is similar. The Limit Comparison Test states that if we have two positive functions f(x) and g(x), and the limit as x approaches a certain value (say, 'c') of f(x)/g(x) is a finite positive number, then the integrals of f(x) and g(x) either both converge or both diverge.

In our case, let's consider f(θ) = tan(4θ) and try to find a suitable g(θ) near θ = π/8. As we discussed, tan(4θ) behaves like 1/cos(4θ) near π/8. So, let’s take g(θ) = 1/cos(4θ). Now, we'll compute the limit:

lim (θ→π/8-) [tan(4θ) / (1/cos(4θ))] = lim (θ→π/8-) [sin(4θ) / cos(4θ) * cos(4θ)] = lim (θ→π/8-) sin(4θ) = sin(π/2) = 1

Since the limit is a finite positive number (1), the integrals of tan(4θ) and 1/cos(4θ) either both converge or both diverge. Now, we need to analyze the integral of 1/cos(4θ) from 0 to π/8. But even this integral is not straightforward. We might need to use another comparison or a different technique.

Let's try a different approach with the Limit Comparison Test. We know that as θ approaches π/8, 4θ approaches π/2. We can use the fact that cos(x) ≈ (π/2 - x) as x approaches π/2. So, cos(4θ) ≈ (π/2 - 4θ) as θ approaches π/8. This suggests we compare with g(θ) = 1/(π/2 - 4θ). Now, let’s look at the limit:

lim (θ→π/8-) [tan(4θ) / (1/(π/2 - 4θ))] = lim (θ→π/8-) [sin(4θ) / cos(4θ) * (π/2 - 4θ)]

This limit is a bit tricky. We can use L'Hôpital's Rule here. First, rewrite the limit as:

lim (θ→π/8-) [sin(4θ) * (π/2 - 4θ) / cos(4θ)]

Applying L'Hôpital's Rule, we differentiate the numerator and denominator:

Numerator derivative: 4cos(4θ)(π/2 - 4θ) - 4sin(4θ) Denominator derivative: -4sin(4θ)

So, the limit becomes:

lim (θ→π/8-) [4cos(4θ)(π/2 - 4θ) - 4sin(4θ)] / [-4sin(4θ)] = lim (θ→π/8-) [cos(4θ)(π/2 - 4θ) - sin(4θ)] / [-sin(4θ)]

As θ approaches π/8, this becomes [0 - 1] / [-1] = 1. The limit is 1, which is finite and positive. So, tan(4θ) behaves like 1/(π/2 - 4θ) near π/8. Now, we consider the integral:

∫[0, π/8] 1/(π/2 - 4θ) dθ

Evaluating the Comparison Integral

Let's evaluate the integral of our comparison function, g(θ) = 1/(π/2 - 4θ), from 0 to π/8:

∫[0, π/8] 1/(π/2 - 4θ) dθ

We can use a simple u-substitution. Let u = π/2 - 4θ, so du = -4 dθ. Then, dθ = -du/4. When θ = 0, u = π/2, and when θ = π/8, u = 0. Our integral becomes:

∫[π/2, 0] (1/u) (-du/4) = (-1/4) ∫[π/2, 0] (1/u) du = (1/4) ∫[0, π/2] (1/u) du

This integral is a classic example of a divergent improper integral. We know that ∫(1/u) du = ln|u|, so we have:

(1/4) [ln|u|] from 0 to π/2

However, we need to be careful because ln(0) is undefined. We need to take a limit:

(1/4) lim (a→0+) [ln(π/2) - ln(a)]

As a approaches 0 from the positive side, ln(a) approaches -∞. Therefore, the limit is:

(1/4) [ln(π/2) - (-∞)] = ∞

This integral diverges! Since the integral of our comparison function diverges, and we showed earlier using the Limit Comparison Test that tan(4θ) behaves like 1/(π/2 - 4θ) near π/8, we can conclude that the original integral also diverges.

Conclusion

So, guys, after carefully analyzing the integral ∫[0, π/8] tan(4θ) dθ using the Direct Comparison Test (and ultimately, the Limit Comparison Test for a cleaner approach), we've determined that this integral diverges. We identified the discontinuity at θ = π/8, explored potential comparison functions, and evaluated the integral of our chosen comparison function. This journey highlights the power and versatility of comparison tests in determining the convergence or divergence of improper integrals.

I hope this breakdown was helpful! Remember, practice makes perfect, so keep exploring different integrals and comparison techniques. You'll become a calculus whiz in no time! Keep your mathematical curiosity alive, and I'll catch you in the next discussion!