Prove The Integral: ∫₀^(π/2) Log(cos X) Dx = -π/2 Log 2

Hey guys! Today, we're diving into a super interesting integral problem from calculus that many students find challenging yet incredibly rewarding. We're going to prove that the definite integral of log(cos x) from 0 to π/2 equals - (π/2) log 2. This isn't just a cool mathematical fact; it's a journey through some neat calculus techniques, including substitution and properties of logarithms. So, buckle up, grab your favorite beverage, and let's get started!

Introduction to the Integral

Before we jump into the proof, let's make sure we're all on the same page. We're dealing with a definite integral, which means we're finding the area under the curve of the function log(cos x) between the limits 0 and π/2. The natural logarithm, denoted as log, is the logarithm to the base e, where e is approximately 2.71828. Understanding the behavior of both the cosine function and the logarithmic function is crucial for tackling this problem. The cosine function oscillates between -1 and 1, and its values are positive in the interval [0, π/2]. The logarithmic function, on the other hand, is only defined for positive arguments. As x approaches 0, cos x approaches 1, and log(cos x) approaches log(1), which is 0. As x approaches π/2, cos x approaches 0, and log(cos x) approaches negative infinity. This behavior gives us a glimpse into the complexity and elegance of the integral we are about to solve.

Setting Up the Integral

Let's define our integral as I. This will make our explanation cleaner and easier to follow. We have:

I = ∫₀^(π/2) log(cos x) dx

This is the integral we want to evaluate. Our strategy involves a clever use of substitution and trigonometric identities to simplify the integral into a form we can easily solve. This integral appears frequently in advanced calculus and is a classic example of using symmetry and substitution to solve seemingly complex problems. Recognizing such patterns is a key skill in mathematical problem-solving, which helps in tackling similar integrals and proving various mathematical theorems.

The Substitution Trick

The first trick up our sleeve is a substitution. We're going to let x = π/2 - u. This might seem like a random move, but it's a brilliant way to exploit the symmetry of the cosine function. When we substitute, we also need to change the limits of integration. When x = 0, u = π/2, and when x = π/2, u = 0. Also, dx becomes -du. So, our integral transforms to:

I = ∫_(π/2)^0 log(cos(π/2 - u)) (-du)

Now, remember the trigonometric identity cos(π/2 - u) = sin(u). This is a crucial step because it connects the cosine function with the sine function, allowing us to use their symmetrical properties. Also, we can flip the limits of integration by changing the sign of the integral:

I = ∫_0^(π/2) log(sin u) du

Notice that we've ended up with a very similar integral, but with sin u instead of cos x. Since u is just a dummy variable, we can rewrite it as x without changing the value of the integral:

I = ∫_0^(π/2) log(sin x) dx

This substitution technique is a cornerstone of integral calculus, demonstrating how a seemingly complex integral can be simplified by changing variables. This method leverages the fundamental theorem of calculus and the properties of trigonometric functions, making it a powerful tool in solving a wide range of problems.

Combining the Integrals

Now comes the fun part! We have two expressions for I:

I = ∫₀^(π/2) log(cos x) dx

I = ∫_0^(π/2) log(sin x) dx

Let's add these two equations together:

2I = ∫₀^(π/2) log(cos x) dx + ∫_0^(π/2) log(sin x) dx

Since the limits of integration are the same, we can combine the integrals:

2I = ∫₀^(π/2) [log(cos x) + log(sin x)] dx

Using the property of logarithms that log(a) + log(b) = log(ab), we get:

2I = ∫₀^(π/2) log(cos x * sin x) dx

This step beautifully combines the two integrals into a single one, making it easier to work with. The application of logarithmic properties not only simplifies the expression but also reveals a hidden symmetry within the integral, paving the way for the next phase of our solution. This combination is a classic move in integral calculus, demonstrating how algebraic manipulations can transform seemingly intractable integrals into manageable forms.

The Double Angle Identity

Now, we need another trick. Recall the double angle identity for sine: sin(2x) = 2 * sin(x) * cos(x). We can rewrite our integrand to use this identity:

2I = ∫₀^(π/2) log((1/2) * 2 * sin(x) * cos(x)) dx

2I = ∫₀^(π/2) log((1/2) * sin(2x)) dx

Using the logarithmic property log(ab) = log(a) + log(b) again, we split the integral:

2I = ∫₀^(π/2) [log(1/2) + log(sin(2x))] dx

2I = ∫₀^(π/2) log(1/2) dx + ∫_0^(π/2) log(sin(2x)) dx

The introduction of the double angle identity is a masterstroke, transforming the integrand into a form that is much easier to handle. By separating the logarithm of the constant term and the sine function, we can address each part individually, simplifying the overall integration process. This step showcases the power of trigonometric identities in integral calculus, highlighting their role in simplifying complex expressions.

Evaluating the First Integral

The first integral is straightforward:

∫₀^(π/2) log(1/2) dx = log(1/2) * ∫_0^(π/2) dx

= log(1/2) * [x]_0^(π/2)

= log(1/2) * (π/2 - 0)

= (π/2) * log(1/2)

Since log(1/2) = log(2^(-1)) = -log(2), we have:

∫₀^(π/2) log(1/2) dx = -(π/2) log(2)

This integral is relatively simple to evaluate, as the integrand is a constant. It directly applies the fundamental theorem of calculus and the properties of logarithms, reinforcing the importance of these foundational concepts in solving more complex integrals. This part of the solution serves as a solid baseline, allowing us to focus on the remaining integral, which requires further manipulation.

Evaluating the Second Integral

Let's tackle the second integral:

∫_0^(π/2) log(sin(2x)) dx

We'll use another substitution: let v = 2x, so dv = 2 dx, and dx = (1/2) dv. When x = 0, v = 0, and when x = π/2, v = π. The integral becomes:

(1/2) ∫_0^π log(sin v) dv

Now, we can split this integral into two parts using the symmetry of the sine function:

(1/2) [∫_0^(π/2) log(sin v) dv + ∫_(π/2)^π log(sin v) dv]

For the second integral, let w = v - π/2, so v = w + π/2 and dv = dw. When v = π/2, w = 0, and when v = π, w = π/2. Also, sin(v) = sin(w + π/2) = cos(w). So, the second integral becomes:

∫_(π/2)^π log(sin v) dv = ∫_0^(π/2) log(cos w) dw

This is the same as our original integral I! So, we have:

(1/2) [∫_0^(π/2) log(sin v) dv + ∫_0^(π/2) log(cos w) dw] = (1/2) [I + I] = I

Thus, the second integral is equal to (1/2) * 2I = I.

This substitution is another elegant maneuver, simplifying the integral by exploiting the periodic nature of the sine function. By changing variables and leveraging the symmetry around π/2, we transformed the integral into a more recognizable form. This technique demonstrates the power of substitution in handling trigonometric integrals and highlights the importance of recognizing patterns and symmetries in mathematical problem-solving.

Putting It All Together

We had:

2I = ∫₀^(π/2) log(1/2) dx + ∫_0^(π/2) log(sin(2x)) dx

We found:

∫₀^(π/2) log(1/2) dx = -(π/2) log(2)

∫_0^(π/2) log(sin(2x)) dx = I

So:

2I = -(π/2) log(2) + I

Subtracting I from both sides, we get:

I = -(π/2) log(2)

And that's it! We've proven that:

∫₀^(π/2) log(cos x) dx = -(π/2) log(2)

This final assembly is where all the pieces come together, showcasing the elegance and coherence of the solution. By combining the results from the evaluated integrals and simplifying the equation, we arrive at the desired result. This step underscores the importance of methodical problem-solving, where each individual step contributes to the overall solution, leading to a satisfying conclusion.

Conclusion

Isn't that awesome? We've successfully navigated through a challenging integral using a combination of substitution, trigonometric identities, and logarithmic properties. This problem isn't just about finding an answer; it's about appreciating the beauty and interconnectedness of mathematical concepts. By mastering these techniques, you'll be well-equipped to tackle similar problems and deepen your understanding of calculus.

So next time you see an integral that looks daunting, remember this journey. Break it down, look for symmetries, and don't be afraid to try different approaches. Keep practicing, keep exploring, and most importantly, keep enjoying the process. You guys are doing great, and I'll catch you in the next math adventure!