Indefinite Integrals: Step-by-Step Solutions

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Let's dive into solving some indefinite integrals! We've got two problems here that involve different techniques, so let's break them down and tackle them one by one. Whether you're brushing up on your calculus skills or seeing this for the first time, we'll go through it together. So, grab your pencils, and let's get started!

1. Solving ∫(3sin2x + x²)dx

Alright, so our first task is to find the indefinite integral of 3sin2x + x². This might look a bit intimidating at first, but don't worry, we can break it down into smaller, more manageable parts. The key here is to remember the basic rules of integration and how they apply to different functions. We'll tackle the sine function and the polynomial term separately, and then combine the results.

Breaking Down the Integral

The first thing we want to do is split the integral into two separate integrals. This is a handy trick because it allows us to focus on each term individually. So, we can rewrite the original integral like this:

∫(3sin2x + x²)dx = ∫3sin2x dx + ∫x² dx

See? Much less scary already! Now we have two simpler integrals to deal with. Let's start with the first one, which involves the sine function.

Integrating 3sin2x

When we're dealing with sin2x, we need to remember the chain rule in reverse. The integral of sin(kx) is (-1/k)cos(kx) + C, where k is a constant and C is the constant of integration. In our case, we have 3sin2x, so k is 2. Let's go through the steps:

  1. Pull out the constant: 3∫sin2x dx
  2. Integrate sin2x: 3 * (-1/2)cos2x + C
  3. Simplify: -3/2 cos2x + C

So, the indefinite integral of 3sin2x is -3/2 cos2x + C. Easy peasy!

Integrating x²

Now, let's move on to the second part: ∫x² dx. This one is a classic application of the power rule for integration. Remember, the power rule states that ∫xⁿ dx = (xⁿ⁺¹)/(n+1) + C, where n is any real number except -1. In our case, n is 2.

Let's apply the power rule:

∫x² dx = (x²⁺¹)/(2+1) + C

Simplify:

∫x² dx = (x³)/3 + C

Voila! The indefinite integral of x² is (x³)/3 + C.

Combining the Results

Now that we've found the indefinite integrals of both parts, all that's left to do is add them together. Remember, each integral has its own constant of integration, but we can combine them into a single constant at the end. So, here's how it looks:

∫(3sin2x + x²)dx = -3/2 cos2x + (x³)/3 + C

And that's it! We've successfully found the indefinite integral of 3sin2x + x². You see, by breaking the problem down and tackling each part separately, it becomes much more manageable. This is a key strategy in calculus, and it's something you'll use again and again.

Key Takeaways for the First Integral

  • Splitting the integral: When you have multiple terms, break them down into separate integrals. This makes each part easier to handle.
  • Sine integral: Remember that the integral of sin(kx) is (-1/k)cos(kx) + C.
  • Power rule: The power rule ∫xⁿ dx = (xⁿ⁺¹)/(n+1) + C is your best friend for polynomial terms.
  • Constant of integration: Don't forget the + C! Indefinite integrals always have a constant of integration.

2. Solving ∫(x + 2)cos3x dx

Okay, let's move on to our second integral: ∫(x + 2)cos3x dx. This one is a bit trickier than the first because it involves the product of two different types of functions: a linear function (x + 2) and a trigonometric function cos3x. To tackle this, we'll need to use a technique called integration by parts. This method is super useful when you have a product of functions inside an integral. Mastering integration by parts is a game-changer in calculus!

Understanding Integration by Parts

Integration by parts is based on the product rule for differentiation. The formula for integration by parts is:

∫u dv = uv - ∫v du

Where:

  • u is a function we choose to differentiate.
  • dv is the remaining part of the integral, which we need to integrate.
  • du is the derivative of u.
  • v is the integral of dv.

The trick is to choose u and dv wisely. We want to pick u such that its derivative, du, is simpler than u, and dv such that it's easy to integrate. Let's see how this works in our case.

Choosing u and dv

In our integral ∫(x + 2)cos3x dx, we need to decide which part will be u and which will be dv. Here's the thought process:

  • If we choose u = (x + 2), then du = dx, which is simpler.
  • If we choose dv = cos3x dx, then v = (1/3)sin3x, which is manageable.

So, it looks like this choice will work well. Let's write it down:

  • u = x + 2
  • dv = cos3x dx

Now, we need to find du and v:

  • du = dx
  • v = ∫cos3x dx = (1/3)sin3x

Applying the Integration by Parts Formula

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

∫u dv = uv - ∫v du

∫(x + 2)cos3x dx = (x + 2)(1/3)sin3x - ∫(1/3)sin3x dx

See how we've transformed the original integral into something that looks a bit easier? The integral on the right side is simpler to solve than the original one. Let's simplify and solve it.

Solving the Remaining Integral

We now have:

∫(x + 2)cos3x dx = (1/3)(x + 2)sin3x - (1/3)∫sin3x dx

We need to find the integral of sin3x. We know that the integral of sin(kx) is (-1/k)cos(kx) + C. So:

∫sin3x dx = (-1/3)cos3x + C

Plugging this back into our equation:

∫(x + 2)cos3x dx = (1/3)(x + 2)sin3x - (1/3)(-1/3)cos3x + C

Simplifying the Result

Let's simplify the expression to get our final answer:

∫(x + 2)cos3x dx = (1/3)(x + 2)sin3x + (1/9)cos3x + C

And there you have it! We've successfully solved the second indefinite integral using integration by parts. It might seem like a lot of steps, but with practice, it becomes second nature. The key is to break it down and follow the formula.

Key Takeaways for the Second Integral

  • Integration by parts: Use this technique when you have a product of functions inside an integral.
  • Choosing u and dv: Pick u such that its derivative is simpler, and dv such that it's easy to integrate.
  • Applying the formula: Remember ∫u dv = uv - ∫v du.
  • Simplifying: Don't forget to simplify your answer after applying the formula.

Final Thoughts

So, guys, we've tackled two pretty interesting indefinite integrals today! We've seen how to break down integrals with multiple terms and how to use integration by parts when we have products of functions. These are fundamental techniques in calculus, and they'll come in handy in all sorts of situations. The most important thing is to practice, practice, practice! The more you work through problems, the more comfortable you'll become with these concepts. Keep up the great work, and happy integrating!