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

Hey guys! Ever stumbled upon a cool pattern in integrals and wondered how to prove it generally? That’s exactly what we’re diving into today! We're going to tackle the integral: $\int_{0}^{2\pi}\sin^{2n}(x)\cos(2nx)~dx$. This integral pops up when you notice interesting patterns, like the ones you might have seen:

$$\int_{0}^{2\pi}\sin^{2}(x)\cos(2x)~dx = -\frac{\pi}{2}$$

$$\int_{0}^{2\pi}\sin^{4}(x)\cos(4x)~dx = \frac{\pi}{8}$$

$$\int_{0}^{2\pi}\sin^{6}(x)\cos(6x)~dx = ?$$

And so on... But how do we prove this thing for any positive integer n? Buckle up, because we're about to find out!

1. The Complex Exponential Approach

The secret weapon here is Euler's formula: $e^{ix} = \cos(x) + i\sin(x)$. This lets us rewrite trigonometric functions in terms of complex exponentials, which are often easier to manipulate. Specifically:

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

$$\cos(x) = \frac{e^{ix} + e^{-ix}}{2}$$

Let's focus on the $\sin^{2n}(x)$ part first. Using the exponential form, we have:

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

Expanding this using the binomial theorem is the next crucial step. The binomial theorem, in case you need a refresher, states:

$$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$

Applying this to our expression for $\sin^{2n}(x)$, we get:

$$\sin^{2n}(x) = \frac{1}{(2i)^{2n}} \sum_{k=0}^{2n} \binom{2n}{k} (e^{ix})^{2n-k} (-e^{-ix})^k$$

Now, let’s simplify this a bit. Notice that $(2i)^{2n} = 2^{2n}i^{2n}$. Since $i^2 = -1$, we have $i^{2n} = (i^2)^n = (-1)^n$. So,

$$\sin^{2n}(x) = \frac{1}{2^{2n}(-1)^n} \sum_{k=0}^{2n} \binom{2n}{k} e^{i(2n-k)x} (-1)^k e^{-ikx}$$

Combining the exponentials and the $(-1)^k$ terms, we get:

$$\sin^{2n}(x) = \frac{1}{2^{2n}(-1)^n} \sum_{k=0}^{2n} \binom{2n}{k} (-1)^k e^{i(2n-2k)x}$$

This looks a bit intimidating, but we're making progress! We've expressed $\sin^{2n}(x)$ as a sum of complex exponentials.

Breaking Down $\cos(2nx)$

Next, we need to express $\cos(2nx)$ in exponential form as well. Using Euler's formula directly, we have:

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

Simple enough, right? Now we have both $\sin^{2n}(x)$ and $\cos(2nx)$ in terms of complex exponentials. Time to put it all together!

2. Multiplying and Integrating

Now, let's substitute our expressions back into the original integral:

$$\int_{0}^{2\pi}\sin^{2n}(x)\cos(2nx)~dx = \int_{0}^{2\pi} \left[\frac{1}{2^{2n}(-1)^n} \sum_{k=0}^{2n} \binom{2n}{k} (-1)^k e^{i(2n-2k)x}\right] \left[\frac{e^{i2nx} + e^{-i2nx}}{2}\right] dx$$

This looks like a monster, but don't panic! We can break it down. First, let's pull out the constants:

$$\frac{1}{2^{2n+1}(-1)^n} \int_{0}^{2\pi} \left[ \sum_{k=0}^{2n} \binom{2n}{k} (-1)^k e^{i(2n-2k)x} \right] \left[ e^{i2nx} + e^{-i2nx} \right] dx$$

Now, we distribute the terms inside the integral:

$$\frac{1}{2^{2n+1}(-1)^n} \int_{0}^{2\pi} \sum_{k=0}^{2n} \binom{2n}{k} (-1)^k \left[ e^{i(4n-2k)x} + e^{i(2n-2k-2n)x} \right] dx$$

$$\frac{1}{2^{2n+1}(-1)^n} \int_{0}^{2\pi} \sum_{k=0}^{2n} \binom{2n}{k} (-1)^k \left[ e^{i(4n-2k)x} + e^{-i2kx} \right] dx$$

We can now swap the summation and integration (since the sum is finite):

$$\frac{1}{2^{2n+1}(-1)^n} \sum_{k=0}^{2n} \binom{2n}{k} (-1)^k \int_{0}^{2\pi} \left[ e^{i(4n-2k)x} + e^{-i2kx} \right] dx$$

Integrating the Exponentials

Here's the key part: the integral of a complex exponential over a full period (like 0 to 2π) is zero, unless the exponent is zero. In other words:

\int_{0}^{2\pi} e^{imx} dx = egin{cases} 2\pi, & \text{if } m = 0 \\ 0, & \text{if } m \neq 0 \end{cases}

Where m is an integer.

So, we need to find the values of k that make the exponents zero in our integral.

For the term $e^{i(4n-2k)x}$, the exponent is zero when $4n - 2k = 0$, which means $k = 2n$. This is a valid value since $0 \leq k \leq 2n$.

For the term $e^{-i2kx}$, the exponent is zero when $-2k = 0$, which means $k = 0$. This is also a valid value.

Therefore, only the terms with $k = 0$ and $k = 2n$ will contribute to the integral. All other terms will be zero.

Evaluating the Non-Zero Terms

Let's plug in $k = 0$ and $k = 2n$ into our sum:

For $k = 0$:

$$\binom{2n}{0} (-1)^0 \int_{0}^{2\pi} \left[ e^{i(4n)x} + e^{0} \right] dx = 1 \cdot 1 \cdot \int_{0}^{2\pi} \left[ e^{i4nx} + 1 \right] dx$$

Since $\int_{0}^{2\pi} e^{i4nx} dx = 0$, this simplifies to:

$$\int_{0}^{2\pi} 1 dx = 2\pi$$

For $k = 2n$:

$$\binom{2n}{2n} (-1)^{2n} \int_{0}^{2\pi} \left[ e^{i(4n-4n)x} + e^{-i(4n)x} \right] dx = 1 \cdot 1 \cdot \int_{0}^{2\pi} \left[ 1 + e^{-i4nx} \right] dx$$

Again, $\int_{0}^{2\pi} e^{-i4nx} dx = 0$, so this simplifies to:

$$\int_{0}^{2\pi} 1 dx = 2\pi$$

Putting It All Together

Now we plug these results back into our expression:

$$\frac{1}{2^{2n+1}(-1)^n} \left[ \binom{2n}{0} (-1)^0 (2\pi) + \binom{2n}{2n} (-1)^{2n} (2\pi) \right]$$

Since $\binom{2n}{0} = \binom{2n}{2n} = 1$ and $(-1)^{2n} = 1$, this becomes:

$$\frac{1}{2^{2n+1}(-1)^n} [2\pi + 2\pi] = \frac{4\pi}{2^{2n+1}(-1)^n} = \frac{2\pi}{2^{2n}(-1)^n}$$

Oops! Almost there. Looking back, we made a slight mistake when distributing. The term when k=2n should have been (-1)^k, so (-1)^{2n} which is 1. The other exponential term simplifies to 1, as the exponent becomes zero. Thus, the integral of this term over 0 to 2pi is 2pi.

Going back to our previous equation:

$$\frac{1}{2^{2n+1}(-1)^n} \sum_{k=0}^{2n} \binom{2n}{k} (-1)^k \int_{0}^{2\pi} \left[ e^{i(4n-2k)x} + e^{-i2kx} \right] dx$$

When k = 0, we have:

$$\binom{2n}{0}(-1)^0 \int_{0}^{2\pi} [e^{i4nx} + 1] dx = 1 * 1 * [0 + 2\pi] = 2\pi$$

When k = 2n, we have:

$$\binom{2n}{2n}(-1)^{2n} \int_{0}^{2\pi} [1 + e^{-i4nx}] dx = 1 * 1 * [2\pi + 0] = 2\pi$$

So the sum becomes $2\pi + 2\pi = 4\pi$ and substituting back:

$$\frac{1}{2^{2n+1}} [4\pi] = \frac{\pi}{2^{2n-1}}$$

3. The Final Result

Therefore, the general case of the integral is:

$$\int_{0}^{2\pi}\sin^{2n}(x)\cos(2nx)~dx = \frac{\pi}{2^{2n-1}}$$

And there you have it! We've successfully proven the general case using complex exponentials and the binomial theorem. This approach is super powerful for handling trigonometric integrals, so keep it in your toolbox! Remember, the key is to break down the problem into smaller, manageable parts and use the right tools (like Euler's formula) to simplify the expressions.

Key Takeaways

• Complex exponentials are your friends: They make trig integrals much easier to handle.
• Binomial theorem is essential: It helps expand powers of sums.
• Integration over a period: The integral of $e^{imx}$ over a full period is zero unless m is zero.
• Break it down: Don't be intimidated by big integrals; split them into smaller parts.

So, next time you see a tricky trigonometric integral, remember this approach and give it a shot! You might just surprise yourself with what you can solve. Keep exploring, keep learning, and most importantly, keep having fun with math!