Integral Of Cos 2x Sinx

Solving the Integral of cos(2x)sin(x): A full breakdown

The integral of cos(2x)sin(x), ∫cos(2x)sin(x)dx, presents a seemingly straightforward challenge in calculus. That said, its solution beautifully illustrates the application of trigonometric identities and integration techniques. Here's the thing — this complete walkthrough will walk you through the process, explaining the underlying principles and providing a deep understanding of the solution. We'll explore multiple approaches, catering to different levels of mathematical understanding. This detailed explanation will equip you with not just the answer, but also the tools to tackle similar integration problems confidently.

I. Introduction: Understanding the Problem

Before diving into the solution, let's understand the problem itself. We are tasked with finding the antiderivative of the function f(x) = cos(2x)sin(x). Consider this: this means finding a function F(x) such that F'(x) = cos(2x)sin(x). In practice, the key to solving this integral lies in manipulating the trigonometric expression using appropriate identities to simplify it into a form that's easily integrable. This process often involves using the product-to-sum formulas, which are invaluable tools in simplifying trigonometric expressions involving products of sine and cosine functions.

II. Method 1: Using the Product-to-Sum Formula

The most straightforward approach involves utilizing the product-to-sum trigonometric identity. This identity allows us to convert the product of two trigonometric functions into a sum or difference of trigonometric functions, making integration significantly simpler. The relevant identity is:

cos(A)sin(B) = ½[sin(A+B) – sin(A-B)]

In our case, A = 2x and B = x. Substituting these values into the identity, we get:

cos(2x)sin(x) = ½[sin(2x+x) – sin(2x-x)] = ½[sin(3x) – sin(x)]

Now, the integral becomes much more manageable:

∫cos(2x)sin(x)dx = ½∫[sin(3x) – sin(x)]dx

We can now integrate term by term:

∫sin(3x)dx = -⅓cos(3x) + C₁ (where C₁ is the constant of integration)

∫sin(x)dx = -cos(x) + C₂ (where C₂ is the constant of integration)

Combining these results, we have:

∫cos(2x)sin(x)dx = ½[-⅓cos(3x) + C₁ + cos(x) - C₂]

Since C₁ and C₂ are arbitrary constants, their difference is also an arbitrary constant. Let's denote this combined constant as 'C':

∫cos(2x)sin(x)dx = -⅓cos(3x) + ½cos(x) + C

This is the final solution obtained using the product-to-sum formula. This method is generally preferred due to its elegance and direct application of a standard trigonometric identity.

III. Method 2: Using Double Angle Formula and Integration by Parts

Alternatively, we can solve this integral using a combination of the double-angle formula for cosine and integration by parts. The double-angle formula for cosine is:

cos(2x) = 1 - 2sin²(x)

Substituting this into our original integral:

∫cos(2x)sin(x)dx = ∫(1 - 2sin²(x))sin(x)dx = ∫[sin(x) - 2sin³(x)]dx

This integral can now be solved by splitting it into two separate integrals:

∫sin(x)dx = -cos(x) + C₁

The second integral, ∫2sin³(x)dx, requires a bit more work. We can use the reduction formula for the integral of sinⁿ(x):

∫sinⁿ(x)dx = - (1/n)sinⁿ⁻¹(x)cos(x) + (n-1)/n ∫sinⁿ⁻²(x)dx

Applying this for n=3:

∫sin³(x)dx = -(1/3)sin²(x)cos(x) + (2/3)∫sin(x)dx = -(1/3)sin²(x)cos(x) - (2/3)cos(x) + C₂

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Substituting these results back into our expression:

∫cos(2x)sin(x)dx = -cos(x) + 2[(1/3)sin²(x)cos(x) + (2/3)cos(x)] + C

Simplifying this expression (remembering that sin²(x) = 1 - cos²(x)), we eventually arrive at the same solution as Method 1:

∫cos(2x)sin(x)dx = -⅓cos(3x) + ½cos(x) + C

While this method is longer, it provides valuable practice with integration by parts and trigonometric reduction formulas.

IV. Explanation of the Steps: A Deeper Dive

Let's break down some key steps in both methods for a more thorough understanding.

Method 1: Product-to-Sum Formula

The crucial step here is applying the product-to-sum formula correctly. Understanding why this formula works involves visualizing the trigonometric functions on the unit circle and using their geometric relationships. The formula arises from the addition and subtraction formulas for sine and cosine.

Method 2: Double Angle Formula and Integration by Parts

Here, the key lies in recognizing that the double angle formula can simplify the expression, and understanding when to apply integration by parts. Integration by parts is a powerful technique used when integrating products of functions, where one function can be easily differentiated and the other easily integrated. The formula for integration by parts is:

∫u dv = uv - ∫v du

Choosing the correct 'u' and 'dv' is vital for successful application. In this case, when dealing with ∫sin³(x)dx, we could use reduction formulas or alternative methods involving trigonometric substitutions to solve this integral.

V. Frequently Asked Questions (FAQ)

• Q: Can other trigonometric identities be used to solve this integral?

A: Yes, although the product-to-sum formula is the most efficient, other identities could be used but might lead to more complicated intermediate steps. The choice often depends on the solver's familiarity with various identities and their preferences.

• Q: Why is the constant of integration 'C' added?

A: The constant of integration 'C' accounts for the fact that the derivative of a constant is zero. So, many functions can have the same derivative. The 'C' represents this family of functions.

• Q: How can I verify my solution?

A: The simplest way to verify your solution is to differentiate it. If the derivative of your solution equals the original integrand (cos(2x)sin(x)), then your solution is correct.

• Q: What if the integral had different angles, for example, ∫cos(3x)sin(2x)dx?

A: The same principles apply. You would use the appropriate product-to-sum formula: cos(A)sin(B) = ½[sin(A+B) - sin(A-B)], substituting A = 3x and B = 2x, then proceed with the integration.

VI. Conclusion: Mastering Trigonometric Integrals

Solving the integral of cos(2x)sin(x) is a valuable exercise that reinforces understanding of trigonometric identities and integration techniques. Mastering these techniques is crucial for tackling more complex integration problems in calculus and related fields like physics and engineering. We've explored two distinct methods, highlighting the versatility of mathematical tools. That said, the more you work through similar problems, the more confident and proficient you'll become in solving trigonometric integrals. Remember that practice is key. Don't hesitate to explore different approaches and choose the method that best suits your understanding and problem-solving style. The journey of understanding calculus is a rewarding one; embrace the challenge and enjoy the process of learning!