Problem 76 - Calculus Early Transcendentals

Question

Evaluate the following integrals. Consider completing the square. $$\int \frac{d x}{\sqrt{(x-1)(3-x)}}$$

Step-by-Step Solution

Answer

Answer: The result of the integral is $\sinh^{-1}(x - 2) + C$.

Step 1: Completing the square

Let's complete the square for the expression in the square root: $$\sqrt{(x-1)(3-x)} = \sqrt{-x^2+4x-3}$$ We can complete the square for the quadratic part by finding the square of half of the coefficient of the x term, i.e., \((4/2)^2 = 4\), then adding and subtracting this value to the expression: $$\sqrt{-x^2 + 4x - 3} = \sqrt{-(x^2 - 4x + 4 - 4 - 3)} = \sqrt{-(x - 2)^2 + 1}$$ Now the integral becomes: $$\int \frac{d x}{\sqrt{-(x - 2)^2 + 1}}$$

Step 2: Substituting for a simpler expression

Let's use the substitution \(x - 2 = \sinh{u}\), with \(u = \sinh^{-1}(x - 2)\). The derivative of x with respect to u is: $$\frac{d x}{d u} = \cosh{u}$$ Applying this substitution to the integral, we get: $$\int \frac{\cosh{u}}{\sqrt{-\sinh^2{u} + 1}} d u$$ We note that \(\cosh^2{u} - \sinh^2{u} = 1.\) Therefore, the integral simplifies to: $$\int \frac{\cosh{u}}{\sqrt{\cosh^2{u} - \sinh^2{u}}} d u$$

Step 3: Final simplification and integration

The expression in the square root simplifies to: $$\int \frac{\cosh{u}}{\cosh{u}} d u = \int d u$$ Now we can integrate directly with respect to \(u\): $$\int d u = u + C$$

Step 4: Replacing the substitution

We now replace \(u\) with the original substitution, which is \(u = \sinh^{-1}(x - 2)\): $$u + C = \sinh^{-1}(x - 2) + C$$ This is the result of the integral: $$\int \frac{d x}{\sqrt{(x-1)(3-x)}} = \sinh^{-1}(x - 2) + C$$

Key Concepts

Completing the Square

Completing the square is a mathematical technique used to transform quadratic expressions into a perfect square added to or subtracted from a constant. This method is particularly useful when evaluating integrals involving square roots of quadratic expressions, as it simplifies the expression under the square root, making the integral more manageable.

For example, consider the quadratic part of the integrand in our exercise, \( -x^2 + 4x - 3 \). To complete the square, we take the coefficient of \( x \), which is 4, divide it by 2, and square the result to get 4. We then add and subtract 4 inside the expression to avoid changing its value, obtaining \( -(x^2 - 4x + 4) + 1 \). This simplifies to \( -(x - 2)^2 + 1 \), a much simpler form that reveals the structure of a hyperbolic function.

Integral Substitution

Integral substitution, often known as 'u-substitution', is a technique in calculus used to find the antiderivatives of complex functions. By changing variables, it transforms a given integral into a simpler, more easily integrable form.

In our case, after completing the square, we make a substitution that relates the expression under the square root to a known identity involving hyperbolic functions. Specifically, we set \( x - 2 = \sinh(u) \), which gives us a new differential relationship: \( \frac{dx}{du} = \cosh(u) \). This substitution simplifies the integral significantly because of the identity \( \cosh^2(u) - \sinh^2(u) = 1 \), which is reminiscent of the Pythagorean identity with trigonometric functions but applied to their hyperbolic counterparts.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions that arise in areas such as geometry, calculus, and complex analysis. The primary functions are the hyperbolic sine (\sinh) and cosine (\cosh), which satisfy the hyperbolic identity \( \cosh^2(u) - \sinh^2(u) = 1 \), resembling the trigonometric Pythagorean identity. These functions prove instrumental when dealing with integrals that contain square roots of quadratic expressions, as seen in our integral.

The use of hyperbolic functions in the substitution step simplifies the integral to an expression involving a quotient of \( \cosh(u) \) and its square root. The identity allows us to cancel the hyperbolic sine and cosine terms, resulting in a simple integral in terms of \( u \), which can then be integrated to find the antiderivative.