Problem 58 - Essential Calculus

Question

Consider the integral \(\int \frac{1}{\sqrt{6 x-x^{2}}} d x\). (a) Find the integral by completing the square of the radicand. (b) Find the integral by making the substitution \(u=\sqrt{x}\). (c) The antiderivatives in parts (a) and (b) appear to be significantly different. Use a graphing utility to graph each antiderivative in the same viewing window and determine the relationship between them. Find the domain of each.

Step-by-Step Solution

Answer

The antiderivatives found in parts (a) and (b) are the same, and each has the domain \([0,6]\). Thus, the integral of \(\int \frac{1}{\sqrt{6 x-x^{2}}} d x\) is \(\arcsin\left(\frac{x - 3}{3}\right) + C\) or \(\arcsin\left(\frac{\sqrt{x}}{\sqrt{6}}\right) + C\).

Step 1: Complete the square

To complete the square for \(6x - x^2\), notice that this can be rewritten as \(- (x^2 - 6x)\). Completing the square inside the parentheses gives us \[-(x^2 - 6x) = -(x - 3)^2 + 9.\] This allows us to simplify the integral to \(\int \frac{1}{\sqrt{-(x-3)^2 + 9}} dx\).

Step 2: Evaluate the integral (part a)

The integral with the completed square is a standard form that can be evaluated using a standard formula for an integral involving \(\sqrt{a^2 - u^2}\): The integral is \(\arcsin\left(\frac{x - 3}{3}\right) + C\), where C is the integration constant.

Step 3: Apply substitution (part b)

Make the substitution \(u = \sqrt{x}\), then \(x = u^2\) and \(dx = 2u du\). The given integral thus simplifies to \(\int \frac{2u}{\sqrt{6u^2 - u^4}} du\), simplify to get \(\int \frac{2}{\sqrt{6 - u^2}} du\). This integral can be solved using the same formula used in step 2 which gives \(\arcsin\left(\frac{u}{\sqrt{6}}\right) + C\). Substituting back in \( u = \sqrt{x} \) gives the final result as \(\arcsin\left(\frac{\sqrt{x}}{\sqrt{6}}\right) + C\).

Step 4: Compare antiderivatives (part c)

Though the two provided answers may look different, they are actually the same mathematical object. This can be shown by graphing both antiderivatives and observing that the graphs overlay each other precisely in the domain of \(x\) from 0 to 6 (except at the points where the function is undefined due to division by zero). Moreover, graphing can reveal that both functions have the same domain, which is \([0,6]\).