Solving ∫₀^(1/2) Arctan(x² - X + 1) Dx Step-by-Step

Introduction

The fascinating world of calculus often presents us with intricate integrals that require a blend of techniques and insights to solve. In this article, we will embark on a journey to decode the definite integral I = ∫₀^(1/2) arctan(x² - x + 1) dx and demonstrate that it elegantly evaluates to ln(2)/2. This particular integral, sourced from the Integration Bee, serves as a quintessential example of how trigonometric identities, algebraic manipulations, and integration strategies converge to unlock a seemingly complex problem. Our exploration will not only provide a step-by-step solution but also delve into the underlying concepts and methodologies, enriching your understanding of integral calculus.

The Challenge: Unveiling the Integral

The problem at hand is to evaluate the definite integral:

I = ∫₀^(1/2) arctan(x² - x + 1) dx

At first glance, the integrand, arctan(x² - x + 1), appears daunting. The quadratic expression within the arctangent function suggests that a direct approach might be cumbersome. Therefore, we will need to employ strategic techniques to simplify the integral and make it tractable. Our journey begins with a clever application of trigonometric identities and a keen observation of the integrand's structure.

Initial Maneuvers: Trigonometric Identities

As hinted in the initial attempt, a crucial step involves leveraging the identity:

arctan(x) = π/2 - arccot(x)

Applying this identity to our integral, we transform it into a more manageable form. This transformation allows us to shift our focus from the arctangent function to the arccotangent, which often simplifies the algebraic manipulations required.

Thus, we rewrite the integral as:

I = ∫₀^(1/2) [π/2 - arccot(x² - x + 1)] dx

This seemingly simple step is pivotal as it opens up new avenues for simplification and evaluation.

Deconstructing the Integrand: Algebraic Insights

Now, let's focus on the arccotangent term. We have arccot(x² - x + 1). The quadratic expression x² - x + 1 can be rewritten to reveal a structure that is amenable to further simplification. By completing the square, we can express the quadratic in a form that highlights its key characteristics.

Completing the square, we have:

x² - x + 1 = (x - 1/2)² + 3/4

This form is particularly insightful because it shows that the quadratic is always positive, which is consistent with the domain of the arccotangent function. Furthermore, it suggests that we might be able to use trigonometric substitutions or identities related to the arccotangent function to further simplify the integral.

A Strategic Substitution: Tangent Subtraction Formula

The expression (x - 1/2)² + 3/4 hints at the possibility of using a tangent subtraction formula. Recall that:

tan(A - B) = (tan(A) - tan(B)) / (1 + tan(A)tan(B))

The goal here is to manipulate the argument of the arccotangent function, x² - x + 1, into a form that resembles the denominator of the tangent subtraction formula. This approach is strategic because it allows us to break down the complex arccotangent term into simpler, more manageable components.

To this end, we can rewrite x² - x + 1 as:

x² - x + 1 = 1 + x(x - 1)

This form is crucial because it now resembles the denominator of the tangent subtraction formula. By carefully choosing tan(A) and tan(B), we can make the argument of the arccotangent function match the desired form.

Let's consider the expression:

(x - (x - 1)) / (1 + x(x - 1)) = 1 / (x² - x + 1)

This expression is significant because it represents the tangent of a difference. Specifically, if we let A = arctan(x) and B = arctan(x - 1), then:

tan(A - B) = tan(arctan(x) - arctan(x - 1)) = (x - (x - 1)) / (1 + x(x - 1)) = 1 / (x² - x + 1)

Taking the arccotangent of both sides, we get:

arccot(x² - x + 1) = arctan(1 / (x² - x + 1)) = arctan(tan(arctan(x) - arctan(x - 1)))

However, since we are dealing with arccotangent, it's more appropriate to write:

arccot(x² - x + 1) = arctan(x) - arctan(x - 1)

This decomposition is a pivotal step in simplifying the integral. We have successfully expressed the complex arccotangent term as a difference of two arctangent functions.

Reassembling the Integral: A Simplified Form

Now that we have decomposed the arccotangent term, we can rewrite the integral I as:

I = ∫₀^(1/2) [π/2 - (arctan(x) - arctan(x - 1))] dx

This form is significantly simpler than the original integral. We can now break it down into individual integrals and evaluate them separately.

I = ∫₀^(1/2) (π/2) dx - ∫₀^(1/2) arctan(x) dx + ∫₀^(1/2) arctan(x - 1) dx

Let's denote these integrals as:

I₁ = ∫₀^(1/2) (π/2) dx

I₂ = ∫₀^(1/2) arctan(x) dx

I₃ = ∫₀^(1/2) arctan(x - 1) dx

Thus, I = I₁ - I₂ + I₃.

Evaluating I₁: A Straightforward Integral

The integral I₁ is straightforward to evaluate:

I₁ = ∫₀^(1/2) (π/2) dx = (π/2) [x]₀^(1/2) = (π/2) (1/2 - 0) = π/4

Evaluating I₂: Integration by Parts

To evaluate I₂, we will use integration by parts. Recall the integration by parts formula:

∫ u dv = uv - ∫ v du

Let u = arctan(x) and dv = dx. Then, du = dx / (1 + x²) and v = x.

Applying integration by parts, we get:

I₂ = ∫₀^(1/2) arctan(x) dx = [x arctan(x)]₀^(1/2) - ∫₀^(1/2) x / (1 + x²) dx

Now, let's evaluate the terms:

[x arctan(x)]₀^(1/2) = (1/2) arctan(1/2) - 0 arctan(0) = (1/2) arctan(1/2)

For the remaining integral, we can use a simple u-substitution. Let u = 1 + x², then du = 2x dx, and x dx = (1/2) du.

Thus, the integral becomes:

∫₀^(1/2) x / (1 + x²) dx = (1/2) ∫(1 to 5/4) du / u = (1/2) [ln|u|]₁^(5/4) = (1/2) [ln(5/4) - ln(1)] = (1/2) ln(5/4)

Therefore,

I₂ = (1/2) arctan(1/2) - (1/2) ln(5/4)

Evaluating I₃: A Shifted Arctangent

Now, let's tackle I₃:

I₃ = ∫₀^(1/2) arctan(x - 1) dx

We can use a simple substitution to make this integral more manageable. Let y = x - 1, then dy = dx. When x = 0, y = -1, and when x = 1/2, y = -1/2.

Thus, the integral becomes:

I₃ = ∫₋₁^(-1/2) arctan(y) dy

Notice that this integral is similar to I₂, but with different limits of integration. We can use integration by parts again. Let u = arctan(y) and dv = dy. Then, du = dy / (1 + y²) and v = y.

Applying integration by parts, we get:

I₃ = [y arctan(y)]₋₁^(-1/2) - ∫₋₁^(-1/2) y / (1 + y²) dy

Now, let's evaluate the terms:

[y arctan(y)]₋₁^(-1/2) = (-1/2) arctan(-1/2) - (-1) arctan(-1) = (-1/2) (-arctan(1/2)) + arctan(-1) = (1/2) arctan(1/2) - π/4

For the remaining integral, we can use a similar u-substitution as before. Let u = 1 + y², then du = 2y dy, and y dy = (1/2) du.

Thus, the integral becomes:

∫₋₁^(-1/2) y / (1 + y²) dy = (1/2) ∫(2 to 5/4) du / u = (1/2) [ln|u|]₂^(5/4) = (1/2) [ln(5/4) - ln(2)]

Therefore,

I₃ = (1/2) arctan(1/2) - π/4 - (1/2) [ln(5/4) - ln(2)]

The Grand Finale: Combining the Results

Now we have evaluated all three integrals: I₁, I₂, and I₃. Recall that:

I = I₁ - I₂ + I₃

Substituting the values we found:

I = π/4 - [(1/2) arctan(1/2) - (1/2) ln(5/4)] + [(1/2) arctan(1/2) - π/4 - (1/2) ln(5/4) + (1/2) ln(2)]

Simplifying the expression, we observe that several terms cancel out:

I = π/4 - (1/2) arctan(1/2) + (1/2) ln(5/4) + (1/2) arctan(1/2) - π/4 - (1/2) ln(5/4) + (1/2) ln(2)

I = (1/2) ln(2)

Thus, we have successfully demonstrated that:

I = ∫₀^(1/2) arctan(x² - x + 1) dx = ln(2) / 2

Conclusion

In this comprehensive exploration, we have meticulously evaluated the definite integral ∫₀^(1/2) arctan(x² - x + 1) dx and shown that it equals ln(2)/2. Our journey involved a strategic blend of trigonometric identities, algebraic manipulations, and integration techniques. We began by transforming the integral using the identity arctan(x) = π/2 - arccot(x). We then cleverly rewrote the quadratic expression within the arccotangent function and employed the tangent subtraction formula to decompose the integrand into simpler terms. Finally, we utilized integration by parts and u-substitution to evaluate the resulting integrals and arrive at the elegant solution. This exercise underscores the power and versatility of calculus in solving intricate problems and highlights the importance of strategic problem-solving approaches.

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