Evaluating ∫tan(x)√(1+sin(x)) Dx A Step By Step Solution

Trigonometric integrals, a cornerstone of calculus, often present intriguing challenges and demand a blend of trigonometric identities, algebraic manipulation, and clever substitutions for successful resolution. In this article, we delve into the evaluation of a specific trigonometric integral, providing a detailed, step-by-step solution and highlighting key techniques applicable to a broader range of similar problems. Our focus will be on the integral ∫tan(x)√(1+sin(x)) dx, a classic example that beautifully illustrates the power of strategic substitutions and trigonometric transformations.

This discussion will take you through a meticulous breakdown of the solution, emphasizing the underlying logic and mathematical principles involved. We'll begin by introducing the problem, outlining the initial approach, and then meticulously working through the necessary substitutions and simplifications. The goal is not just to arrive at the final answer but to provide a clear, pedagogical explanation that enhances understanding and problem-solving skills in the realm of trigonometric integrals. So, let's embark on this mathematical journey and unravel the intricacies of this integral.

Our objective is to evaluate the indefinite integral ∫tan(x)√(1+sin(x)) dx. This integral brings together trigonometric functions in a form that isn't immediately amenable to standard integration techniques. The presence of both the tangent function and the square root term suggests that a clever substitution might be the key to simplification. Our initial strategy involves exploring substitutions that can help us eliminate the square root and express the integrand in a more manageable form. A natural starting point is to consider the expression inside the square root, 1+sin(x), and explore whether a substitution involving this term can lead to progress.

Given the structure of the integral, a promising approach is to use a substitution that directly addresses the square root term. By setting 1+sin(x) = t², we aim to transform the square root into a simple variable, t. This substitution, however, necessitates expressing the entire integral in terms of t, which requires careful consideration of the other trigonometric functions present, namely tan(x). The relationship between sin(x) and cos(x), and how they relate to tan(x), will play a crucial role in this transformation. The initial substitution also brings in the differential dx, which needs to be expressed in terms of dt using the derivative of our substitution. This step is vital in ensuring the integral is entirely in terms of the new variable, t, allowing us to apply standard integration methods.

Let's proceed with the substitution 1+sin(x) = t². This implies that sin(x) = t² - 1. To find the differential relationship, we differentiate both sides with respect to x, which gives us cos(x) dx = 2t dt. From this, we can express dx as dx = (2t dt) / cos(x). Now, we need to express cos(x) in terms of t. Recall the fundamental trigonometric identity sin²(x) + cos²(x) = 1. Substituting sin(x) = t² - 1, we get (t² - 1)² + cos²(x) = 1. Solving for cos²(x) gives us cos²(x) = 1 - (t² - 1)² = 1 - (t⁴ - 2t² + 1) = -t⁴ + 2t². Thus, cos(x) = √(-t⁴ + 2t²) = t√(2 - t²).

Now we can express dx completely in terms of t: dx = (2t dt) / (t√(2 - t²)) = (2 dt) / √(2 - t²). Next, we need to address tan(x). Recall that tan(x) = sin(x) / cos(x). We already have sin(x) = t² - 1 and cos(x) = t√(2 - t²). Therefore, tan(x) = (t² - 1) / (t√(2 - t²)). Now we can substitute everything back into the original integral: ∫tan(x)√(1+sin(x)) dx = ∫((t² - 1) / (t√(2 - t²))) * t * (2 dt) / √(2 - t²) = ∫2(t² - 1) / (2 - t²) dt. This simplifies to ∫2(t² - 1) / (2 - t²) dt. This transformed integral, while still requiring some manipulation, is significantly more approachable than the original one. The absence of square roots and the presence of a rational function allows us to employ techniques such as partial fraction decomposition or algebraic manipulation to further simplify and integrate.

Having transformed the original integral into ∫2(t² - 1) / (2 - t²) dt, our next step is to simplify and integrate this rational function. A common technique for integrating rational functions is to perform polynomial long division or algebraic manipulation to express the integrand in a more manageable form. In this case, we can rewrite the integrand as follows: 2(t² - 1) / (2 - t²) = -2(1 - t²) / (2 - t²) = -2 * ((2 - t²) - 1) / (2 - t²) = -2 * (1 - 1/(2 - t²)) = -2 + 2/(2 - t²). This manipulation separates the integrand into a constant term and a simpler rational function, making it easier to integrate.

Now our integral becomes ∫(-2 + 2/(2 - t²)) dt. We can split this into two separate integrals: ∫-2 dt + ∫2/(2 - t²) dt. The first integral, ∫-2 dt, is straightforward and equals -2t + C₁. The second integral, ∫2/(2 - t²) dt, requires a bit more attention. To evaluate it, we can use partial fraction decomposition or recognize that it is related to a standard integral form. We can rewrite 2 - t² as (√2 - t)(√2 + t). Using partial fraction decomposition, we can express 2/(2 - t²) as A/(√2 - t) + B/(√2 + t) for some constants A and B. Solving for A and B, we find that A = 1/√2 and B = 1/√2. Thus, our integral becomes ∫(1/√2)/(√2 - t) dt + ∫(1/√2)/(√2 + t) dt.

These integrals are now in a form that we can readily integrate. The integral ∫(1/√2)/(√2 - t) dt evaluates to -(1/√2)ln|√2 - t| + C₂ and the integral ∫(1/√2)/(√2 + t) dt evaluates to (1/√2)ln|√2 + t| + C₃. Combining these results, we have ∫2/(2 - t²) dt = (1/√2)(ln|√2 + t| - ln|√2 - t|) + C₄. Using the logarithm property ln(a) - ln(b) = ln(a/b), we can further simplify this to (1/√2)ln|(√2 + t)/(√2 - t)| + C₄. Finally, combining the two integrals, we have ∫(-2 + 2/(2 - t²)) dt = -2t + (1/√2)ln|(√2 + t)/(√2 - t)| + C, where C is the constant of integration.

Having evaluated the integral in terms of t, our final step is to substitute back in terms of x to obtain the result in the original variable. Recall that we made the substitution 1+sin(x) = t², so t = √(1+sin(x)). Substituting this back into our result, we get: -2√(1+sin(x)) + (1/√2)ln|(√2 + √(1+sin(x)))/(√2 - √(1+sin(x)))| + C. This expression represents the indefinite integral of tan(x)√(1+sin(x)) with respect to x.

While this result may appear complex, it is the culmination of a series of strategic substitutions, algebraic manipulations, and the application of integration techniques. The key to solving this integral lies in the initial substitution, which simplifies the square root term and allows us to express the integrand in a more manageable form. The subsequent steps involve careful simplification, partial fraction decomposition, and the use of standard integral forms. The final back-substitution returns the result to the original variable, providing the complete solution to the problem.

In summary, the integral ∫tan(x)√(1+sin(x)) dx evaluates to -2√(1+sin(x)) + (1/√2)ln|(√2 + √(1+sin(x)))/(√2 - √(1+sin(x)))| + C. This example showcases the power of substitution and algebraic manipulation in solving trigonometric integrals. The process highlights the importance of identifying appropriate substitutions, simplifying the integrand, and applying known integration techniques to arrive at the final solution. This comprehensive approach not only solves the specific problem but also provides valuable insights and skills applicable to a wide range of integration challenges.

In this exploration, we have successfully evaluated the integral ∫tan(x)√(1+sin(x)) dx, navigating through the complexities of trigonometric integrals with strategic substitutions and algebraic manipulations. The initial move of setting 1+sin(x) = t² proved crucial, transforming the integral into a more tractable form. This substitution allowed us to eliminate the square root and express the integrand as a rational function in terms of t. The subsequent steps involved simplifying the rational function, employing partial fraction decomposition, and integrating the resulting terms.

The process underscored the importance of recognizing appropriate substitutions in tackling challenging integrals. The ability to transform an integral into a simpler form is a fundamental skill in calculus, often requiring a blend of intuition and experience. Furthermore, the use of trigonometric identities and algebraic manipulations played a significant role in simplifying the integrand and making it amenable to integration techniques. The final back-substitution ensured that the result was expressed in terms of the original variable, completing the solution.

This example serves as a valuable case study for students and practitioners of calculus, illustrating the power and versatility of integration techniques. The detailed step-by-step solution provides a clear roadmap for approaching similar problems, emphasizing the importance of careful planning, strategic execution, and thorough simplification. The journey through this integral highlights the beauty and elegance of calculus, where seemingly complex problems can be unraveled with the right tools and techniques. The result, while intricate, stands as a testament to the power of mathematical reasoning and the satisfaction of solving a challenging problem.

In conclusion, the evaluation of ∫tan(x)√(1+sin(x)) dx offers a rich learning experience, reinforcing key concepts in calculus and demonstrating the art of problem-solving in the context of trigonometric integrals. The skills and insights gained from this exploration can be applied to a wide range of mathematical challenges, empowering individuals to tackle complex problems with confidence and creativity.