Problem 59 - Calculus Single Variable

Question

Calculate the given integral. $$ \int \frac{1}{\sqrt{2 x-x^{2}}} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \arcsin(x-1) + C \).

Step 1: Identify the Integral

Recognize that the integral is of the form \( \int \frac{1}{\sqrt{2x-x^2}} \ dx \). This suggests using trigonometric substitution to simplify the integrand.

Step 2: Complete the Square

Rewrite \( 2x - x^2 \) by completing the square: \( 2x - x^2 = -(x^2 - 2x) = -(x^2 - 2x + 1 - 1) = -(x-1)^2 + 1 \). Thus, the integrand becomes \( \int \frac{1}{\sqrt{1-(x-1)^2}} \ dx \).

Step 3: Utilize a Trigonometric Substitution

Let \( x-1 = \sin \theta \), so \( dx = \cos \theta \ d\theta \). This transforms the integral into \( \int \frac{1}{\cos \theta} \cos \theta \ d\theta \), which simplifies to \( \int d\theta \).

Step 4: Integrate w.r.t. Theta

Integrate \( \int d\theta \) easily to get \( \theta + C \).

Step 5: Back-Substitute for x

Replace \( \theta \) with the expression in terms of \( x \). Since \( x-1 = \sin \theta \), \( \theta = \arcsin(x-1) \). Thus, the solution becomes \( \arcsin(x-1) + C \).

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique in calculus used to simplify integrals involving square roots. It's particularly effective when dealing with expressions resembling the forms of Pythagorean identities, such as \(\sqrt{a^2 - x^2}\).

In our problem, after completing the square, we re-write the expression under the square root as \(1 - (x-1)^2\). Recognizing this form allows us to use the substitution \(x-1 = \sin \theta\). Here’s why this works:

• The Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\) guarantees that the square root simplifies nicely, turning it into a simple expression involving cosine.
• This substitution transforms the integral into one with respect to \(\theta\), often simplifying the integral considerably.
• By substituting back after integrating, we can express the solution in terms of the original variable \(x\).

Trigonometric substitution is particularly useful because it often transforms a complex integral into a much simpler one that's easy to solve.

Completing the Square

Completing the square is a crucial algebraic method used to manipulate quadratic expressions into a form that is easier to work with. This technique is especially handy for integrals involving quadratic polynomials.

In our integral, the expression under the square root is initially \(2x - x^2\). Completing the square involves the following steps:

• We first rearrange terms to get \(-x^2 + 2x\). Notice the negative, initially outside the parenthesis.
• Next, express this as \(-(x^2 - 2x)\) and adjust this expression by adding and subtracting 1 to complete the square: this gives \(-(x-1)^2 + 1\).
• This new form is more recognizable and easier to work with, allowing us to apply trigonometric substitution efficiently.

Completing the square adjusts the quadratic into a perfect square trinomial, making the problem more approachable and solvable.

Integral Calculus

Integral calculus is the branch of calculus focused on integrals and their applications, such as area, accumulation, and near-limit calculations. Integrals can be broadly divided into two types: definite and indefinite.

An integral is designed to find the total accumulation of quantities, such as area under a curve or volume of a shape. In this problem, \(\int \frac{1}{\sqrt{2x-x^2}} \, dx\), we are dealing with an indefinite integral, meaning there are no limits of integration provided.

Here, the primary challenge involves simplifying the expression under the integral sign to facilitate integration. Methods like completing the square and trigonometric substitution help us transition from a complicated integral into a simpler form that is directly solvable.

The process typically involves strategy: identify the form, manipulate the expression for simplicity (often using algebraic methods), choose suitable substitutions, and then compute the integral using direct anti-differentiation techniques.

Definite and Indefinite Integrals

Integrals can be categorized into definite and indefinite integrals, each serving distinct purposes in calculus.

• Definite integrals calculate the actual numerical value of area or accumulation over a specified interval. These involve upper and lower limits of integration, representing the interval over which you wish to calculate.
• Indefinite integrals, like \(\int \frac{1}{\sqrt{2x-x^2}} \, dx\), provide a general form or family of functions that represent accumulations or anti-derivatives without specific limits. These include the constant of integration \(C\), reflecting the set of all possible solutions.

Understanding the distinction between these two is crucial as it influences how one approaches and interprets mathematical problems. In real-world applications, definite integrals often occur when precision and specific intervals are required, such as in physics or economics, whereas indefinite integrals lay the groundwork for solving differential equations and algebraic integration techniques.