91Ó°ÊÓ

Trigonometric identities are essential tools in calculus and algebra for transforming complex expressions into simpler ones. While this specific exercise does not directly involve trigonometric identities like sine, cosine, or tangent, it helps to know that substitutive methods often work similarly, utilizing these identities to make expressions more manageable. In substitution for integrals, the goal is often to use a relationship or identity to transform an integral into a simpler form that is recognizable and easier to evaluate. When problems do require trigonometric identities, these can include:

• Pythagorean Identities: Such as \( \sin^2\theta + \cos^2\theta = 1 \).
• Angle Sum and Difference: Like \( \sin(a \pm b) = \sin a \cos b \pm \cos a \sin b \).
• Double Angle: For example, \( \sin 2a = 2 \sin a \cos a \).

These identities simplify integrals, especially those involving periodic functions, allowing practitioners to transform integrals into more familiar standard forms.

Algebraic methods help in solving integrals by transforming the integrand through mathematical manipulation. The exercise uses substitution, an algebraic technique that simplifies the integral by changing variables, effectively solving it in fewer steps. When we see an expression like \( x - \sqrt{x} \), using substitution with \( u = \sqrt{x} \) is strategic because it creates a simpler polynomial. Setting \( u = \sqrt{x} \) implies \( x = u^2 \), allowing us to replace \( x - \sqrt{x} \) with \( u^2 - u \). Differential substitution is kept in line by calculating \( dx = 2u \, du \). Additionally, algebraic manipulation often involves factoring expressions, as seen in \( u^2 - u = u(u - 1) \). This process can lead to a canceled term in the integrand, simplifying the integral further. Once the substitution and algebraic simplifications are completed, the resulting expression is sometimes easier to integrate, as simpler integrals or standard forms such as \( \ln|a| \) arise.

Evaluating an integral involves calculating an expression's antiderivative. This exercise used substitution to reduce the integral's complexity. After substituting \( u = \sqrt{x} \) and differentiating, we replace the integral in terms of \( u \). With substitution complete, the integral \( \int \frac{2u \, du}{u(u-1)} \) is simplified by canceling \( u \) from the numerator and denominator, resulting in \( 2 \int \frac{du}{u-1} \). This clearly aligns with a standard integral form, \( \int \frac{du}{a + u} \), which typically evaluates to \( \ln|u - a| \). Integrating \( 2 \frac{du}{u-1} \) gives \( 2 \ln|u - 1| + C \), where \( C \) represents the constant of integration. The last crucial step involves back-substituting to the original variable \( x \) using the relationship \( u = \sqrt{x} \), leading to the solution expression \( 2 \ln|\sqrt{x} - 1| + C \).This evaluation method showcases how integrals can be made straightforward through tactical substitutions and simplifications, producing solutions more readily interpretable in contexts where functions are analytically investigated.