91Ó°ÊÓ

Completing the square is a powerful algebraic technique that helps simplify quadratic expressions. This method transforms a standard quadratic equation of the form \(ax^2 + bx + c\) into a perfect square trinomial, making it easier to integrate, especially in cases involving definite integrals.

In the given exercise, the expression in the denominator, \(2x^2 - 2x + 1\), can be simplified using completing the square. First, we factor out the coefficient of \(x^2\), which is 2, giving us \(2(x^2 - x + \frac{1}{2})\).

To complete the square, we focus on \(x^2 - x + \frac{1}{2}\). Take the coefficient of \(x\), which is -1, halve it and square it: \(\left(-\frac{1}{2}\right)^2 = \frac{1}{4}\). Adding and subtracting this value within the expression, we obtain a perfect square trinomial: \((x - \frac{1}{2})^2 + \frac{1}{4}\).

This allows us to rewrite the denominator as \(2((x - \frac{1}{2})^2 + \frac{1}{4})\), which is a vital step for solving the integral using other methods such as substitution.

The substitution method is a strategic approach to simplify and evaluate integrals. By changing variables, we can transform a complicated integral into a simpler one. This technique is particularly useful when dealing with expressions that resemble a standard form or require a transformation for easier integration.

In our problem, after completing the square in the denominator, we introduce a substitution where \(u = x - \frac{1}{2}\). This simplifies the integral by shifting the limits of integration and aligning our expression with standard integral forms.

With this substitution:

• \(dx = du\), which maintains the differential unchanged.
• The limits of integration change as well. When \(x = 0\), \(u = -\frac{1}{2}\); and when \(x = 1\), \(u = \frac{1}{2}\).

Substituting these into the integral, we now have the expression \(\frac{1}{2} \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{1}{u^2 + \left(\frac{1}{2}\right)^2} \, du\). This substitution aligns the integral into a form conducive to employing the arctangent function, which is crucial to evaluating the definite integral effectively.

The arctangent function, \(\tan^{-1}(x)\), is the inverse of the tangent function and plays a central role in evaluating certain types of integrals, especially those that can be expressed in the form \(\int \frac{1}{a^2 + x^2} \, dx\).

In this exercise, after using substitution, the integral presents itself as \(\frac{1}{2} \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{1}{u^2 + (\frac{1}{2})^2} \, du\), which fits perfectly with the standard form where \(a^2 = (\frac{1}{2})^2\).
For integrals of this form, the solution involves \(\frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right)\), evaluated across the new limits.

Here, \(a = \frac{1}{2}\), so the integral simplifies to:

• \(\left[ \tan^{-1}\left(\frac{u}{\frac{1}{2}}\right) \right]_{-\frac{1}{2}}^{\frac{1}{2}}\)
• This resolves to \(\tan^{-1}(1) - \tan^{-1}(-1)\).
• Knowing \(\tan^{-1}(1) = \frac{\pi}{4}\) and \(\tan^{-1}(-1) = -\frac{\pi}{4}\), the expression becomes \(\frac{\pi}{4} - (-\frac{\pi}{4}) = \frac{\pi}{2}\).

Thus, the utility of the arctangent function in this context is crucial for reaching the final evaluated solution of the integral.