Problem 32 - Calculus of a Single Variable

Question

Evaluating a Definite Integral In Exercises \(21-32\) evaluate the definite integral. $$ \int_{0}^{1 / \sqrt{2}} \frac{\arccos x}{\sqrt{1-x^{2}}} d x $$

Step-by-Step Solution

Answer

The value of the given definite integral is \(\pi^2 / 16\)

Step 1: Identify the derivative of \(\arccos x\)

The derivative of \(\arccos x\) is \(-\frac{1}{\sqrt{1-x^2}}\). Notice that if we multiply it by \(-1\), it appears in the integrand. This connects with the integral of the form \(\int f'(x) * f(x) dx\), which can be solved by a substitution method.

Step 2: Substitute

Set \(u = \arccos x\). Then \(du = -\frac{1}{\sqrt{1-x^2}} dx\). The original integral can be rewritten in terms of \(u\). Don't forget to change the limits of integral. When \(x = 0\), \(u = \arccos 0 = \frac{\pi}{2}\), and when \(x = 1 / \sqrt{2}\), \(u = \arccos 1 / \sqrt{2} = \frac{\pi}{4}\). So the integral becomes \(-\int_{\pi / 2}^{\pi / 4} u du\).

Step 3: Solve the Integral

Now the integral is straightforward. We just need to integrate \(u\), and then apply the Fundamental Theorem of Calculus to find the definite integral. The antiderivative of \(u\) is \(1/2 * u^2\), so we have \(-[\frac{1}{2} u^2]_{\pi / 2}^{\pi / 4} = -[\frac{1}{2} (\pi / 4)^2 - \frac{1}{2} (\pi / 2)^2]. = -(\pi^2 / 32 - \pi^2 / 8) = \pi^2 / 16\).

Key Concepts

Arccos Function

The arccosine function, often denoted as \( \arccos x \), is the inverse of the cosine function. It takes a value from -1 to 1 and returns an angle ranging from 0 to \( \pi \). This function is important in trigonometry and calculus since many problems involve inverse trigonometric functions.

When dealing with derivatives in calculus, the derivative of \( \arccos x \) is particularly notable. It is given by \(-\frac{1}{\sqrt{1-x^2}}\). This derivative is essential because it simplifies certain integral problems, especially those involving substitution.

For instance, in the original exercise, the expression \( \frac{\arccos x}{\sqrt{1-x^2}} \) uses the derivative of \( \arccos x \) within the integral, facilitating the use of substitution to solve it.

Substitution Method

The substitution method in calculus is a technique used to transform complex integrals into simpler ones. By substituting a part of the integrand with a new variable, the integration process can become more manageable.

In this exercise, we employed the substitution \( u = \arccos x \). This choice is crucial because it aligns with the derivative \(-\frac{1}{\sqrt{1-x^2}}\), which is part of the integrand. The substitution facilitates rewriting the integral in terms of \( u \) instead of \( x \).

During substitution:

• Ensure to replace the differential as well: \( du = -\frac{1}{\sqrt{1-x^2}} dx \).
• Update the limits of integration to match the new variable \( u \). So, \( x = 0 \) changes to \( u = \frac{\pi}{2} \), and \( x = 1 / \sqrt{2} \) to \( u = \frac{\pi}{4} \).

This transforms the integral into a simpler form, \(-\int_{\pi / 2}^{\pi / 4} u\; du\), which can be easily solved.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation and integration, providing a way to evaluate definite integrals. It states that if you have an antiderivative \( F(x) \) of a continuous function \( f(x) \) on an interval \([a, b]\), then the definite integral of \( f \) from \( a \) to \( b \) is \( F(b) - F(a) \).

In the context of the given exercise, after rewriting the integral using substitution, we arrive at an expression that involves an antiderivative of a simpler function: \(-\int_{\pi / 2}^{\pi / 4} u\; du\).

The antiderivative of \( u \) is \( \frac{1}{2}u^2 \). Using the Fundamental Theorem of Calculus, we calculate:

• Plug \( u = \frac{\pi}{4} \) into the antiderivative: \( \frac{1}{2} \left( \frac{\pi}{4} \right)^2 \).
• Plug \( u = \frac{\pi}{2} \) into the antiderivative: \( \frac{1}{2} \left( \frac{\pi}{2} \right)^2 \).

The definite integral then becomes \(- \left( \frac{\pi^2}{32} - \frac{\pi^2}{8} \right)\), resulting in \( \frac{\pi^2}{16} \). This highlights how we use the Fundamental Theorem to find definite integrals efficiently.