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Completing the square is a useful algebraic method used to simplify quadratic expressions, especially when dealing with integrals that include square roots. The goal is to express the quadratic in the form \((x + a)^2 + b\). This can make expressions easier to handle and prepare them for substitution techniques. To complete the square for an expression like \(x^2 + 2x + 5\), follow these steps:

• Identify the linear coefficient: it's 2 in this case.
• Halve this coefficient to get 1, then square it to achieve 1 as well.
• Add and subtract the square (1) within the expression: \(x^2 + 2x + 1 - 1 + 5\).
• Rearrange to achieve a perfect square: \((x+1)^2 + 4\).

This final form \((x+1)^2 + 4\) will help in applying trigonometric substitution, which simplifies the integration process further.

Trigonometric substitution is a powerful technique used in calculus to simplify integrals, particularly those involving square roots of quadratic expressions. It's especially handy when the expression matches patterns like \(a^2 + x^2\), \(x^2 - a^2\), or \(a^2 - x^2\). For example, in our integral, the form is \((x+1)^2 + 4\), suggesting a substitution related to \(\tan(\theta)\) or \(\sec(\theta)\). Consider the substitution \(x + 1 = 2\tan(\theta)\):

• Use this substitution to rewrite the expression: \((x+1)^2 + 4 = 4\sec^2(\theta)\)
• Differentiate \(x+1 = 2\tan(\theta)\) to find \(dx = 2\sec^2(\theta)\,d\theta\)

The original integral transforms to an expression in terms of \(\theta\), which is generally easier to integrate using standard trigonometric identities.The power of trigonometric substitution lies in its ability to convert complex square root expressions into simpler forms that are manageable with standard integration methods.

Integral calculus focuses on finding the integral of functions, which is the process of determining the area under a curve. In this exercise, we've combined completing the square with trigonometric substitution to transform the integral into a form that can be easily evaluated. Once the integral is set up with our substitution, we proceed to integrate each term separately.In the solution, after simplifying, we had to find:

• \(\int 6\tan(\theta)\sec(\theta)\, d\theta\)
• \(\int -3\sec(\theta)\, d\theta\)

Each of these can be integrated using known identities:

• The integral of \(\tan(\theta)\sec(\theta)\) is \(\sec(\theta)+C\).
• The integral of \(-\sec(\theta)\) involves a logarithmic function: \(-3\ln |\sec(\theta) + \tan(\theta)| + C\).

Substituting back the trigonometric identities in terms of \(x\) gives the solution to the original problem, demonstrating the essence and beauty of integral calculus.