Problem 67 - Thomas Calculus in SI Units

Question

Evaluate the integrals. $$\int \frac{d x}{\sqrt{-x^{2}+4 x-3}}$$

Step-by-Step Solution

Answer

\(\int \frac{dx}{\sqrt{-x^2+4x-3}} = \arcsin(x-2) + C\).

Step 1: Complete the Square

To simplify the expression under the square root, we first complete the square for \(-x^2 + 4x - 3\). Start by factoring out the negative sign: \\(- (x^2 - 4x) - 3\). Now, complete the square inside the parentheses: \\(- (x^2 - 4x + 4 - 4) - 3 = -((x-2)^2 - 4) - 3\). So, it simplifies to \(-(x-2)^2 + 4 - 3 = -(x-2)^2 + 1\). The integral becomes \(\int \frac{dx}{\sqrt{1 - (x-2)^2}}\).

Step 2: Use Trigonometric Substitution

Notice that \(\sqrt{1 - (x-2)^2}\) resembles the identity \(\sqrt{1-u^2}\) which can be solved using the substitution \(u = \sin(\theta)\). Let \(x-2 = \sin(\theta)\). Then \(dx = \cos(\theta) d\theta\) and \(\sqrt{1-(x-2)^2} = \sqrt{1-\sin^2(\theta)} = \cos(\theta)\). The integral becomes \(\int \frac{\cos(\theta) d\theta}{\cos(\theta)}\), which simplifies to \(\int d\theta\).

Step 3: Integrate and Back-Substitute

Integrate with respect to \(\theta\): \(\int d\theta = \theta + C\). Since \(\sin(\theta) = x-2\), find \(\theta\) using \(\theta = \arcsin(x-2)\). Therefore, the integral evaluates to \(\arcsin(x-2) + C\).

Step 4: Provide the Final Answer

Thus, the integral \(\int \frac{dx}{\sqrt{-x^2+4x-3}}\) equals \(\arcsin(x-2) + C\), where \(C\) is the constant of integration.

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a clever technique used to simplify integrals that involve square roots and quadratic expressions. This technique is very useful when you see an expression resembling Pythagorean identities, such as \(1 - u^2\). In this exercise, we substitute with the identity \(u = \sin(\theta)\), which reduces the square root into a more manageable form.

The key steps include:

• Identifying the expression under the square root.
• Selecting a substitution based on trigonometric identities, like \(\sin^2(\theta) + \cos^2(\theta) = 1\).

Performing trigonometric substitution allows us to express the integral in terms of \(\theta\), leading to simpler elementary integrals like \(\int d\theta\). Once integrated, we reverse the substitution to express the result back in terms of the original variable, \(x\).

This substitution method, though initially appearing complex, efficiently turns a tough integral into an easily solvable one with just basic trigonometric integration.

Completing the Square

Completing the square is a method used to simplify quadratic expressions, which is particularly useful in integration involving square roots. When faced with a quadratic entity within an integral, it often helps to "complete the square" to transform the expression into a form \((x-h)^2+k\), which is easier to handle.

Here's how it works:

• Start by factoring out any common terms, typically focusing on the coefficient of \(x^2\).
• Rearrange and find the perfect square trinomial by adding and subtracting the square of half the linear coefficient.

This transformation allows the expression under the square root to be rewritten in a simpler, more manageable form. Following this, the integral can often be tackled using trigonometric substitution or other integration methods. Especially for radicals, this approach reveals hidden patterns and identities that simplify calculations considerably.

Through this method, an otherwise cumbersome expression becomes succinct, enabling further techniques to be applied seamlessly.

Definite and Indefinite Integrals

In calculus, integrals can be classified as definite or indefinite based on their purpose and formulation. An indefinite integral refers to the integral of a function with respect to a variable, producing a family of functions represented by \(F(x) + C\), where \(C\) is the constant of integration. This type encapsulates the concept of antiderivatives and reflects a broad set of solutions.

On the other hand, a definite integral computes the area under the curve of a function between two specific limits \([a, b]\). This type results in a numerical value, rather than an algebraic expression. It involves evaluating the antiderivative at these boundaries, \(F(b) - F(a)\).

The choice between definite and indefinite integrals comes down to the problem context:

• Indefinite integrals are used when you need a general solution or antiderivative.
• Definite integrals provide specific values, often related to physical quantities like area, volume, or cumulative change.

Understanding the distinction and applications of both types of integral is fundamental in solving real-world problems or working through complex mathematical expressions.