Evaluating The Integral Of Sin(x)cos(2x) From Π/4 To 3π/4

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Introduction

In this article, we will delve into the process of evaluating the definite integral of the function sin(x)cos(2x) over the interval [π/4, 3π/4]. This type of problem frequently appears in calculus courses and requires a solid understanding of trigonometric identities and integration techniques. Mastering these skills is crucial for anyone pursuing studies in mathematics, physics, engineering, or related fields. This exploration not only provides a step-by-step solution but also underscores the underlying principles and strategies involved in solving definite integrals, particularly those involving trigonometric functions. By carefully dissecting the problem, we aim to provide clarity and enhance your problem-solving abilities in calculus. Throughout this discourse, the strategic application of trigonometric identities to simplify the integrand will be highlighted, alongside the methodical use of integration techniques to arrive at a precise solution. This endeavor will not only demonstrate how to compute such integrals but also illuminate the broader landscape of mathematical problem-solving, emphasizing the importance of strategic thinking and the application of fundamental principles. The detailed approach outlined herein will serve as a valuable resource for students and enthusiasts alike, offering a pathway to deepen their comprehension and proficiency in integral calculus.

Problem Statement

Our primary objective is to calculate the definite integral:

∫π/43π/4sin⁡xcos⁡2x dx\int_{\pi / 4}^{3 \pi / 4} \sin x \cos 2x \, dx

This integral presents a common challenge in calculus, requiring us to integrate a product of trigonometric functions over a specified interval. To effectively tackle this problem, we will employ a strategic combination of trigonometric identities and integration techniques. Initially, we will utilize a trigonometric identity to simplify the integrand, transforming it into a more manageable form that is easier to integrate. Subsequently, we will apply standard integration rules to find the antiderivative of the simplified function. Finally, we will evaluate the antiderivative at the bounds of integration, namely π/4 and 3π/4, to determine the definite integral's value. This process highlights the critical role of both theoretical knowledge and practical skills in calculus, demonstrating how understanding fundamental identities and techniques can lead to efficient and accurate solutions. The integral's specific structure, involving the product of sine and cosine functions with differing arguments, necessitates a thoughtful approach. By breaking down the problem into distinct steps, we aim to demystify the integration process and provide a clear, accessible solution path. This detailed exploration serves not only to solve the given integral but also to reinforce the broader principles of integral calculus and trigonometric manipulation.

Strategy

The key to solving this integral lies in simplifying the integrand, sin(x)cos(2x), using a trigonometric identity. We can rewrite cos(2x) using the double-angle formula:

cos(2x) = cos^2(x) - sin^2(x)$ or $cos(2x) = 1 - 2sin^2(x)$ or $cos(2x) = 2cos^2(x) - 1

The most suitable form for our purpose is cos(2x) = 1 - 2sin²(x). Substituting this into the integral, we get:

∫π/43π/4sin⁡x(1−2sin⁡2x) dx\int_{\pi / 4}^{3 \pi / 4} \sin x (1 - 2\sin^2 x) \, dx

Expanding the integrand gives us:

∫π/43π/4(sin⁡x−2sin⁡3x) dx\int_{\pi / 4}^{3 \pi / 4} (\sin x - 2\sin^3 x) \, dx

Now, we need to address the sin³(x) term. We can rewrite it as sin³(x) = sin(x)sin²(x) and then use the Pythagorean identity sin²(x) = 1 - cos²(x):

sin3(x)=sin(x)(1−cos2(x))sin^3(x) = sin(x)(1 - cos^2(x))

Substituting this back into our integral, we have:

∫π/43π/4(sin⁡x−2sin⁡x(1−cos⁡2x)) dx\int_{\pi / 4}^{3 \pi / 4} (\sin x - 2\sin x(1 - \cos^2 x)) \, dx

This strategic manipulation of trigonometric identities is pivotal in transforming the integral into a form that is more amenable to direct integration. By breaking down complex trigonometric functions into simpler components, we pave the way for the application of standard integration techniques. The choice of the double-angle formula and the subsequent use of the Pythagorean identity are not arbitrary; they are guided by the goal of expressing the integrand in terms of functions whose antiderivatives are well-known. This approach highlights the importance of pattern recognition and strategic thinking in calculus. The ability to identify and apply appropriate identities is a hallmark of a proficient problem-solver in mathematics. The transformation of sin³(x) into a form involving cos²(x) is a crucial step, as it sets the stage for a u-substitution that will simplify the integration process further. This careful orchestration of trigonometric identities and algebraic manipulation underscores the elegance and power of mathematical methods in solving complex problems.

Step-by-Step Solution

  1. Expand the integrand:

∫π/43π/4(sin⁡x−2sin⁡x+2sin⁡xcos⁡2x) dx\int_{\pi / 4}^{3 \pi / 4} (\sin x - 2\sin x + 2\sin x \cos^2 x) \, dx

  1. Simplify:

∫π/43π/4(−sin⁡x+2sin⁡xcos⁡2x) dx\int_{\pi / 4}^{3 \pi / 4} (-\sin x + 2\sin x \cos^2 x) \, dx

  1. Separate the integral:

∫π/43π/4−sin⁡x dx+∫π/43π/42sin⁡xcos⁡2x dx\int_{\pi / 4}^{3 \pi / 4} -\sin x \, dx + \int_{\pi / 4}^{3 \pi / 4} 2\sin x \cos^2 x \, dx

  1. Solve the first integral:

∫π/43π/4−sin⁡x dx=[cos⁡x]π/43π/4=cos⁡(3π/4)−cos⁡(π/4)=−22−22=−2\int_{\pi / 4}^{3 \pi / 4} -\sin x \, dx = [\cos x]_{\pi / 4}^{3 \pi / 4} = \cos(3\pi / 4) - \cos(\pi / 4) = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2}

  1. Solve the second integral using u-substitution:

Let u = cos(x), then du = -sin(x) dx. The limits of integration change as well: when x = π/4, u = cos(π/4) = √2/2, and when x = 3π/4, u = cos(3π/4) = -√2/2.

∫π/43π/42sin⁡xcos⁡2x dx=−2∫2/2−2/2u2 du\int_{\pi / 4}^{3 \pi / 4} 2\sin x \cos^2 x \, dx = -2 \int_{\sqrt{2} / 2}^{-\sqrt{2} / 2} u^2 \, du

=−2[u33]2/2−2/2=−2((−2/2)33−(2/2)33)= -2 \left[ \frac{u^3}{3} \right]_{\sqrt{2} / 2}^{-\sqrt{2} / 2} = -2 \left( \frac{(-\sqrt{2} / 2)^3}{3} - \frac{(\sqrt{2} / 2)^3}{3} \right)

=−2(−2224−2224)=−2(−212−212)=−2(−26)=23= -2 \left( \frac{-2\sqrt{2}}{24} - \frac{2\sqrt{2}}{24} \right) = -2 \left( \frac{-\sqrt{2}}{12} - \frac{\sqrt{2}}{12} \right) = -2 \left( -\frac{\sqrt{2}}{6} \right) = \frac{\sqrt{2}}{3}

  1. Combine the results:

∫π/43π/4sin⁡xcos⁡2x dx=−2+23=−223\int_{\pi / 4}^{3 \pi / 4} \sin x \cos 2x \, dx = -\sqrt{2} + \frac{\sqrt{2}}{3} = -\frac{2\sqrt{2}}{3}

The systematic approach to solving this integral underscores the importance of breaking down complex problems into manageable steps. The initial expansion and simplification of the integrand, facilitated by trigonometric identities, set the stage for direct integration and u-substitution. The separation of the integral into two parts allows for focused attention on each component, minimizing the likelihood of errors. The careful application of the u-substitution technique, including the transformation of integration limits, demonstrates a thorough understanding of calculus principles. The meticulous evaluation of each integral and the subsequent combination of results highlight the importance of precision in mathematical calculations. This step-by-step solution not only provides the final answer but also serves as a pedagogical tool, illustrating the thought process and techniques involved in solving definite integrals. The strategic use of substitutions and the algebraic manipulation of terms underscore the versatility and power of calculus methods in tackling a wide range of mathematical problems. By meticulously following each step, learners can develop a deeper appreciation for the intricacies of integral calculus and enhance their problem-solving skills.

Final Answer

Therefore, the definite integral evaluates to:

∫π/43π/4sin⁡xcos⁡2x dx=−223\int_{\pi / 4}^{3 \pi / 4} \sin x \cos 2x \, dx = -\frac{2\sqrt{2}}{3}

This final result encapsulates the culmination of our step-by-step solution, providing a precise value for the definite integral. The journey from the initial problem statement to this final answer highlights the essential role of strategic simplification, trigonometric identities, and integration techniques in calculus. The negative sign of the result indicates that the net area under the curve of the function sin(x)cos(2x) over the interval [π/4, 3π/4] lies below the x-axis. This interpretation of the integral's value provides a geometric insight into the function's behavior within the specified interval. The magnitude of the result, 2√2/3, quantifies this net area, offering a concrete measure of the function's cumulative effect. The entire process, from problem definition to final answer, exemplifies the rigorous and systematic nature of mathematical problem-solving. The successful evaluation of this definite integral not only provides a numerical solution but also reinforces the underlying principles and techniques of calculus, enhancing the learner's ability to tackle similar challenges in the future. The clarity and precision demonstrated throughout this solution serve as a testament to the power of mathematical methods in unraveling complex problems and arriving at definitive answers.