Solving ∫sin(x)cos(x) Dx A Comprehensive Guide

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Hey guys! Let's dive into the world of integrals and tackle a classic problem that often pops up in calculus: evaluating the integral of sin(x)cos(x). This integral is a fantastic example that showcases the power of different integration techniques, particularly u-substitution. We'll break down the problem step by step, exploring a couple of cool approaches to arrive at the solution. So, grab your pencils, and let's get started!

Understanding the Integral: ∫sin(x)cos(x) dx

Before we jump into solving, let's take a moment to understand what this integral represents. The integral ∫sin(x)cos(x) dx asks us to find a function whose derivative is sin(x)cos(x). In other words, we're looking for the antiderivative of the product of the sine and cosine functions. This type of problem often hints at using a substitution method, as the derivative of one part of the expression might be related to the other part. The core concept revolves around reversing the process of differentiation. When we differentiate a function, we find its rate of change; integration, in essence, is the opposite – we're finding the original function given its rate of change. This is why integration is often called “antidifferentiation.” In our specific case, we have a product of two trigonometric functions, sin(x) and cos(x). This structure suggests that a clever substitution might simplify the integral. The key observation here is that the derivative of sin(x) is cos(x), and the derivative of cos(x) is -sin(x). This relationship between the functions and their derivatives is a strong indicator that u-substitution will be a fruitful approach. Understanding this underlying principle helps us to select the appropriate technique and makes the integration process much more manageable. So, with this in mind, let's move on to exploring how u-substitution can help us crack this integral.

Method 1: U-Substitution with u = sin(x)

The first approach we'll explore involves using u-substitution, a powerful technique for simplifying integrals. The key here is to choose a 'u' that makes the integral easier to handle. In this case, let's try setting u = sin(x). This is a strategic move because the derivative of sin(x) is cos(x), which conveniently appears in our integral. So, if we let u = sin(x), then the next step is to find du, which is the derivative of u with respect to x, multiplied by dx. The derivative of sin(x) is cos(x), so du = cos(x) dx. Now, look back at our original integral: ∫sin(x)cos(x) dx. Notice that we have both sin(x) and cos(x) dx present. We can directly substitute u for sin(x) and du for cos(x) dx. This transforms our integral into a much simpler form: ∫u du. This new integral is straightforward to evaluate. The integral of u with respect to u is simply (1/2)u^2 + C, where C is the constant of integration. Remember, we always add the constant of integration when finding indefinite integrals, as the derivative of a constant is zero. We're not quite done yet, though. We need to express our answer in terms of the original variable, x. Since we set u = sin(x), we substitute sin(x) back in for u. This gives us our final result: (1/2)sin^2(x) + C. So, by using u-substitution with u = sin(x), we've successfully evaluated the integral. This demonstrates how a clever choice of substitution can significantly simplify complex integrals.

Step-by-step Breakdown:

  1. Let u = sin(x)
  2. Then du = cos(x) dx
  3. Substitute: ∫sin(x)cos(x) dx becomes ∫u du
  4. Integrate: ∫u du = (1/2)u^2 + C
  5. Substitute back: (1/2)u^2 + C becomes (1/2)sin^2(x) + C

Method 2: U-Substitution with u = cos(x)

Now, let's switch things up and see what happens if we choose a different substitution. Instead of u = sin(x), let's try setting u = cos(x). This might seem like a similar approach, but it leads to a slightly different path to the solution, highlighting the flexibility of u-substitution. If we let u = cos(x), we need to find du, the derivative of u with respect to x, multiplied by dx. The derivative of cos(x) is -sin(x), so du = -sin(x) dx. Notice that our original integral, ∫sin(x)cos(x) dx, has sin(x) dx, but we have -sin(x) dx in our du. No problem! We can simply multiply both sides of the du equation by -1 to get -du = sin(x) dx. Now we have everything we need for the substitution. We replace cos(x) with u and sin(x) dx with -du. This transforms our integral into ∫u(-du), which simplifies to -∫u du. This integral is just as easy to evaluate as the one we got in the first method. The integral of u with respect to u is (1/2)u^2, so -∫u du = -(1/2)u^2 + C, where C is the constant of integration. Again, we need to express our answer in terms of x. We substitute cos(x) back in for u, giving us -(1/2)cos^2(x) + C. So, by using u-substitution with u = cos(x), we've arrived at another valid solution for the integral. This underscores an important point: sometimes there are multiple ways to solve an integral, and different substitutions can lead to different-looking but equivalent answers. This flexibility is one of the things that makes calculus so interesting and powerful.

Step-by-step Breakdown:

  1. Let u = cos(x)
  2. Then du = -sin(x) dx, so -du = sin(x) dx
  3. Substitute: ∫sin(x)cos(x) dx becomes -∫u du
  4. Integrate: -∫u du = -(1/2)u^2 + C
  5. Substitute back: -(1/2)u^2 + C becomes -(1/2)cos^2(x) + C

Method 3: Using the Trigonometric Identity sin(2x) = 2sin(x)cos(x)

Let's explore a third, and perhaps more elegant, approach to solving this integral: using a trigonometric identity. This method showcases how leveraging trigonometric identities can often simplify integrals involving trigonometric functions. The key identity we'll use here is the double-angle formula for sine: sin(2x) = 2sin(x)cos(x). This identity is a cornerstone of trigonometry and comes in handy in many calculus problems. Looking back at our original integral, ∫sin(x)cos(x) dx, we notice that it closely resembles the right-hand side of the identity. To make it an exact match, we can multiply and divide the integral by 2: ∫sin(x)cos(x) dx = (1/2)∫2sin(x)cos(x) dx. Now, we can directly apply the trigonometric identity, replacing 2sin(x)cos(x) with sin(2x). This transforms our integral into (1/2)∫sin(2x) dx. This new integral is much simpler to handle. To evaluate it, we can use another quick u-substitution. Let's set v = 2x. Then dv = 2 dx, so dx = (1/2)dv. Substituting these into our integral, we get (1/2)∫sin(v) (1/2)dv = (1/4)∫sin(v) dv. The integral of sin(v) with respect to v is -cos(v), so (1/4)∫sin(v) dv = -(1/4)cos(v) + C, where C is the constant of integration. Finally, we substitute back to express our answer in terms of x. Since v = 2x, we get -(1/4)cos(2x) + C. So, by using the trigonometric identity sin(2x) = 2sin(x)cos(x), we've found yet another valid solution for the integral. This method highlights the importance of recognizing trigonometric identities and how they can provide a more direct path to the answer. It also reinforces the idea that integrals can often be solved in multiple ways, each with its own unique elegance.

Step-by-step Breakdown:

  1. Use the identity: sin(2x) = 2sin(x)cos(x)
  2. Rewrite the integral: ∫sin(x)cos(x) dx = (1/2)∫2sin(x)cos(x) dx = (1/2)∫sin(2x) dx
  3. Let v = 2x, then dv = 2 dx, so dx = (1/2)dv
  4. Substitute: (1/2)∫sin(2x) dx becomes (1/4)∫sin(v) dv
  5. Integrate: (1/4)∫sin(v) dv = -(1/4)cos(v) + C
  6. Substitute back: -(1/4)cos(v) + C becomes -(1/4)cos(2x) + C

Comparing the Results: Are They the Same?

Okay, guys, we've tackled this integral using three different methods, and we've arrived at three seemingly different answers: (1/2)sin^2(x) + C, -(1/2)cos^2(x) + C, and -(1/4)cos(2x) + C. This might seem a bit perplexing at first. Are these answers actually the same? The short answer is: yes, they are! But how can this be? Well, the key lies in the constant of integration, C, and trigonometric identities. Remember, the constant of integration means that there's an infinite number of functions that have the same derivative. Our three results are simply different ways of expressing the same family of functions. To see how these answers are equivalent, we can use the Pythagorean trigonometric identity: sin^2(x) + cos^2(x) = 1. Let's start by comparing (1/2)sin^2(x) + C and -(1/2)cos^2(x) + C. We can rewrite (1/2)sin^2(x) as (1/2)(1 - cos^2(x)) using the Pythagorean identity. This gives us (1/2) - (1/2)cos^2(x) + C. Notice that this is the same as -(1/2)cos^2(x) + (1/2 + C). Since C is an arbitrary constant, (1/2 + C) is also an arbitrary constant, so we can simply call it C'. This shows that (1/2)sin^2(x) + C and -(1/2)cos^2(x) + C are indeed equivalent. Now, let's bring in our third result, -(1/4)cos(2x) + C. We can use the double-angle formula for cosine: cos(2x) = cos^2(x) - sin^2(x). Substituting this into our expression, we get -(1/4)(cos^2(x) - sin^2(x)) + C. We can further rewrite this as -(1/4)cos^2(x) + (1/4)sin^2(x) + C. Again, using the Pythagorean identity, we can replace sin^2(x) with (1 - cos^2(x)), giving us -(1/4)cos^2(x) + (1/4)(1 - cos^2(x)) + C, which simplifies to -(1/2)cos^2(x) + (1/4) + C. Once again, we see that this is equivalent to -(1/2)cos^2(x) + C', where C' is a new constant (C + 1/4). So, we've shown that all three answers are equivalent, differing only by a constant. This is a crucial concept in integration and highlights the importance of the constant of integration. It also demonstrates the interconnectedness of trigonometric identities and how they can be used to manipulate and simplify trigonometric expressions. This equivalence is not just a mathematical curiosity; it's a fundamental aspect of integration that reflects the fact that the antiderivative of a function is not unique, but rather a family of functions that differ by a constant.

Conclusion

Alright, guys, we've successfully evaluated the integral of sin(x)cos(x) dx using three different methods: u-substitution with u = sin(x), u-substitution with u = cos(x), and leveraging the trigonometric identity sin(2x) = 2sin(x)cos(x). We've seen how each method provides a unique pathway to the solution, and we've also demonstrated how these seemingly different solutions are actually equivalent, differing only by a constant. This exercise underscores the power and flexibility of integration techniques, as well as the importance of understanding trigonometric identities. Whether you prefer the directness of u-substitution or the elegance of trigonometric manipulation, the key is to choose the method that you find most intuitive and that best suits the problem at hand. Remember, practice makes perfect, so keep exploring different integrals and honing your skills. The world of calculus is full of fascinating challenges, and mastering integration is a crucial step in your mathematical journey. So, keep up the great work, and happy integrating!