Problem 39 - Thomas Calculus

Question

Evaluate each integral in Exercises \(37-42\) by completing the square and using a substitution to reduce it to standard form. $$ \int \frac{d t}{\sqrt{-t^{2}+4 t-3}} $$

Step-by-Step Solution

Answer

The integral evaluates to \( \arcsin(t - 2) + C \).

Step 1: Complete the Square

The expression inside the square root is \(-t^2 + 4t - 3\). We start by completing the square for the quadratic expression. First, take the coefficient of \(t\), which is 4, divide it by 2 to get 2, and then square it to get 4. Now, rewrite the quadratic expression as:\(-t^2 + 4t - 3 = -(t^2 - 4t) - 3 = -(t^2 - 4t + 4 - 4) - 3 = -((t - 2)^2 - 4) - 3= -(t - 2)^2 + 4 - 3= -(t - 2)^2 + 1.\)Thus, the integral becomes:\(\int \frac{dt}{\sqrt{-(t-2)^2 + 1}}\).

Step 2: Simplify the Expression

The expression inside the square root is now \(-(t-2)^2 + 1\).This simplifies to \(1 - (t-2)^2\), which is a form similar to the identity for the sine function, \(1 - x^2\), commonly seen in trigonometric substitution scenarios.

Step 3: Use a Trigonometric Substitution

Use the substitution \(t - 2 = \sin(\theta)\), which implies \(dt = \cos(\theta)d\theta\). Substitute back into the integral:\[\int \frac{dt}{\sqrt{1 - (t-2)^2}} = \int \frac{\cos(\theta)d\theta}{\sqrt{1 - \sin^2(\theta)}}\]Since \(\sqrt{1 - \sin^2(\theta)} = \cos(\theta)\), the integral simplifies to:\[\int d\theta\]

Step 4: Integrate and Substitue Back

The integral \(\int d\theta\) is straightforward:\(\theta + C.\)Since \(\sin(\theta) = t - 2\), we find \(\theta = \arcsin(t - 2)\). Thus, the integral evaluates to:\(\theta + C = \arcsin(t - 2) + C. \)

Key Concepts

Completing the Square

Completing the square is a method used to rewrite a quadratic expression in a way that makes it easier to work with, especially when dealing with integrals or forming perfect squares. In our case, we started with the expression \[-t^2 + 4t - 3\]. This represents a parabola, but we want to transform it into a form that highlights its vertex, which helps in simplification.

• Take the coefficient of the linear term, which is 4 in this expression, and divide by 2 to find the halfway point, giving 2.
• Square this result to get 4.
• Add and subtract this squared term within the quadratic to complete the square.

By completing the square, we change the expression to:\[-(t - 2)^2 + 1\]. Now the quadratic is in perfect square form. It centers the expression around \(t = 2\), making subsequent steps, like trigonometric substitution, easier.

Trigonometric Identities

Trigonometric identities are equations that are true for all values of the variables involved. In integration, these identities can simplify expressions significantly. When we examine our problem, the format we created, \[1 - (t-2)^2\], is reminiscent of the Pythagorean identity: \[\cos^2(\theta) + \sin^2(\theta) = 1\].Understanding this identity allows us to use trigonometric substitution efficiently. The substitution \(t-2 = \sin(\theta)\) exploits this identity:

• By setting \(t-2 = \sin(\theta)\), \(dt\) becomes \(\cos(\theta)d\theta\).
• The expression \(\sqrt{1 - \sin^2(\theta)}\) reduces to \(\cos(\theta)\), aiding in integration.

The power of this approach lies in how well the trigonometric identities simplify the expressions, making integrals more manageable to solve.

Definite and Indefinite Integrals

In calculus, integrals are a fundamental concept with two main types: definite and indefinite integrals. In the context of this exercise, we're dealing with an indefinite integral, \[\int \frac{dt}{\sqrt{-t^2 + 4t - 3}}\].An indefinite integral represents a family of functions and includes a constant of integration, often denoted as \(C\). These types of integrals are used to find antiderivatives.Steps for such integration:

• Complete the square to aid in simplification (as done with \(-t^2 + 4t - 3\)).
• Perform a trigonometric substitution to match the integrand to a known derivative or integral form.
• Integrate the simplified expression, often using basic antiderivative knowledge.
• Substitute back into the original variables and add the constant \(C\).

For this problem, the integration of \(\int d\theta\) yields \(\theta\), which we then substitute back using \(\theta = \arcsin(t - 2)\).This carefully executed process underscores the simplicity and elegance of indefinite integrals when utilizing proper techniques.