Proving The General Case Of Integral ∫₀²π Sin²ⁿ(x)cos(2nx) Dx

Hey guys! So, we've got this cool integral problem we're trying to crack, and it looks like there's a pattern emerging. Let's dive into how we can prove the general case of the integral $ \int_{0}^{2\pi}\sin{2n}(x)\cos(2nx)~dx$. We've already seen some interesting results for specific values of n, and now we want to show it works for any n. Buckle up, because we're about to embark on a calculus adventure!

Unveiling the Pattern: Initial Observations

Before we jump into the nitty-gritty proof, let's recap the pattern we've spotted. We've noticed that:

• When n = 1: $ \int_{0}^{2\pi}\sin{2}(x)\cos(2x)~dx = -\frac{\pi}{2} $
• When n = 2: $ \int_{0}^{2\pi}\sin{4}(x)\cos(4x)~dx = \frac{\pi}{8} $
• When n = 3: (We'll calculate this, but we expect a similar form)

It appears that the integral results in some multiple of $ \pi $, and the coefficient changes with n. Our mission is to figure out the general formula for this coefficient. This is where things get interesting, and we need to pull out some cool calculus tricks to solve it.

To really nail this, we need to dive deep into trigonometric identities and complex numbers. These are the tools that will help us transform this integral into something we can handle. We're not just looking for a quick answer; we want a solid, rock-solid proof that covers all cases. So, let's roll up our sleeves and get to work!

The Strategy: Trigonometric Identities and Complex Numbers

Okay, guys, let's talk strategy. To tackle this integral, we're going to use a powerful combo: trigonometric identities and complex numbers. These are like our secret weapons in the world of calculus, allowing us to transform complicated expressions into more manageable forms. Trust me; it's going to be awesome!

First up, we'll leverage Euler's formula, which connects complex exponentials with trigonometric functions. Remember that gem? It states:

$ e^{ix} = \cos(x) + i\sin(x) $

This formula is a game-changer because it allows us to express sine and cosine in terms of complex exponentials. Why is this cool? Because exponentials are way easier to manipulate than trigonometric functions, especially when we're dealing with powers and products. So, we're going to rewrite $ \sin(x) $ and $ \cos(2nx) $ using Euler's formula. This is our first big step in simplifying the integral. We're essentially translating the problem from the world of trig functions to the world of complex numbers, where the math is often cleaner and more streamlined.

Next, we'll use the binomial theorem to expand the $ \sin^{2n}(x) $ term. This might sound intimidating, but it's just a systematic way of multiplying out the powers of a binomial. By expanding this term, we'll get a sum of complex exponentials, which we can then integrate term by term. This is a classic technique in calculus: break down a complex problem into smaller, more manageable pieces. Each term in the expansion will be simpler to integrate, and we'll be able to see how the pattern emerges.

Finally, we'll integrate each term and then convert back to real numbers. Remember, our original problem was in the real world (no imaginary numbers!), so we need to make sure our final answer is real too. This involves carefully combining terms and using trigonometric identities to simplify the result. It's like solving a puzzle, where each step brings us closer to the final answer. By the end, we'll have a beautiful, elegant formula for the integral, and we'll have proven it works for all values of n. How cool is that?

Step-by-Step Proof: Cracking the Integral

Alright, folks, let's get our hands dirty and dive into the step-by-step proof. We're going to break down this problem into bite-sized pieces, so it's super clear how we arrive at the final answer. Remember, the key here is to use trigonometric identities and complex exponentials to simplify the integral. So, let's start cooking!

Step 1: Expressing Sine and Cosine Using Euler's Formula

First things first, let's rewrite $ \sin(x) $ and $ \cos(2nx) $ in terms of complex exponentials using Euler's formula. As we discussed, Euler's formula is our trusty sidekick in this adventure. We know that:

$ e^{ix} = \cos(x) + i\sin(x) $

and

$ e^{-ix} = \cos(x) - i\sin(x) $

From these, we can express $ \sin(x) $ as:

$ \sin(x) = \frac{e^{ix} - e^{-ix}}{2i} $

and $ \cos(2nx) $ as:

$ \cos(2nx) = \frac{e^{i2nx} + e^{-i2nx}}{2} $

This is a crucial step because it transforms our trigonometric functions into complex exponentials, which are much easier to work with. We've essentially swapped out our old tools for shiny new ones that are better suited for the job.

Step 2: Substituting into the Integral

Now, let's substitute these expressions back into our original integral. This is where the magic really starts to happen. Our integral becomes:

$ \int_{0}^{2\pi} \left(\frac{e^{ix} - e^{{-ix}}{2i}\right)}{2n} \left(\frac{e^{i2nx} + e^{-i2nx}}{2}\right) dx $

Okay, I know what you're thinking: