Evaluate ∫x Sin(x) Dx: Integration By Parts Guide

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Evaluating the Definite Integral of x sin(x) Using Integration by Parts

Hey guys! Today, we're diving into a classic calculus problem: evaluating the definite integral of x sin(x) from 0 to π/2. We're going to tackle this using a powerful technique called integration by parts. This method is super useful when you have a product of two functions inside your integral, like our x and sin(x) here. So, let's get started and break down how to solve this step by step!

Understanding Integration by Parts

Before we jump into the specifics, let's quickly recap the integration by parts formula. It's derived from the product rule for differentiation and looks like this:

∫ u dv = uv - ∫ v du

Where:

  • u is a function we choose to differentiate.
  • dv is the remaining part of the integrand that we choose to integrate.

The key to successfully using integration by parts is choosing the right u and dv. The goal is to pick a u that simplifies when differentiated and a dv that's easy to integrate. This method allows us to transform a complicated integral into something more manageable. In our case, we'll see how this works beautifully.

Setting Up the Problem: Choosing u and dv

Okay, let's apply this to our integral:

∫ 0π/2 x sin(x) dx

We need to decide what to call u and what to call dv. Here's where a little strategy comes in. We want to pick u so that its derivative, du, is simpler than u itself. Looking at our integrand, x becomes 1 when differentiated, which is definitely simpler. On the other hand, sin(x) is easy to integrate. So, let's make these choices:

  • u = x
  • dv = sin(x) dx

Now we need to find du and v. This involves a simple differentiation and integration:

  • du = dx (the derivative of x with respect to x)
  • v = -cos(x) (the integral of sin(x) is -cos(x))

See how choosing u as x made du a simple dx? That's exactly what we wanted!

Applying the Integration by Parts Formula

Now that we have our u, dv, du, and v, we can plug them into the integration by parts formula:

∫ u dv = uv - ∫ v du

Substituting our values, we get:

∫ 0π/2 x sin(x) dx = [x(-cos(x))]0π/2 - ∫ 0π/2 -cos(x) dx

Let's break this down a bit. The first term, [x(-cos(x))]0π/2, comes directly from the uv part of the formula. The second term, - ∫ 0π/2 -cos(x) dx, is the new integral we need to solve.

Notice that the integral on the right side is much simpler than our original integral. That's the magic of integration by parts in action!

Simplifying and Evaluating the New Integral

Let's simplify the expression a bit. We have a negative sign inside the integral, so we can pull that out and make it a positive:

∫ 0π/2 x sin(x) dx = -[xcos(x)]0π/2 + ∫ 0π/2 cos(x) dx

Now, we need to evaluate the definite integral of cos(x) from 0 to π/2. This is a standard integral that we know:

∫ cos(x) dx = sin(x) + C

So, our expression becomes:

∫ 0π/2 x sin(x) dx = -[xcos(x)]0π/2 + [sin(x)]0π/2

Now we're ready to evaluate the terms at the limits of integration.

Evaluating the Definite Integral at the Limits

We need to plug in the upper limit (π/2) and the lower limit (0) into our expression and subtract the results. Let's start with the first term, -[xcos(x)]:

  • At x = π/2: -(π/2)cos(π/2) = -(π/2)(0) = 0
  • At x = 0: -(0)cos(0) = -(0)(1) = 0

So, the first term evaluates to 0 - 0 = 0.

Now let's look at the second term, [sin(x)]:

  • At x = π/2: sin(π/2) = 1
  • At x = 0: sin(0) = 0

So, the second term evaluates to 1 - 0 = 1.

The Final Result

Putting it all together, we have:

∫ 0π/2 x sin(x) dx = 0 + 1 = 1

Therefore, the definite integral of x sin(x) from 0 to π/2 is 1. Awesome, right?

Key Takeaways and Why This Matters

Let's recap what we've done and why it's important:

  • We successfully evaluated the definite integral ∫ 0π/2 x sin(x) dx using integration by parts.
  • The integration by parts formula is ∫ u dv = uv - ∫ v du.
  • Choosing the right u and dv is crucial. We picked u = x because it simplifies when differentiated.
  • We broke down a complex integral into simpler parts, making it solvable.

Why does this matter? Integration by parts is a fundamental technique in calculus and is used extensively in physics, engineering, and other fields. Whenever you encounter an integral with a product of functions, integration by parts is often the key to unlocking the solution. It’s a powerful tool for simplifying integrals that might otherwise seem impossible.

Practice Makes Perfect

The best way to master integration by parts is to practice! Try working through different examples with varying functions. Experiment with different choices for u and dv to see how they affect the difficulty of the integral. You'll start to develop an intuition for which choices lead to simpler integrals.

Here are a few practice problems you can try:

  1. ∫ x cos(x) dx
  2. ∫ x2 ex dx
  3. ∫ ln(x) dx (Hint: think about what you could choose for dv)

Remember, calculus is a journey, not a destination. Keep practicing, keep exploring, and you'll become a master of integration in no time!

Conclusion

So, guys, we've successfully navigated the world of integration by parts and solved a classic definite integral problem. We've seen how choosing the right u and dv can make all the difference, and we've reinforced the importance of this technique in calculus and beyond. Keep practicing, stay curious, and happy integrating!