Problem 95 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(85-94\) $$ \int_{0}^{1} \frac{\tan ^{-1} x}{1+x^{2}} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{\pi^2}{32} \).

Step 1: Understanding the Problem

We need to evaluate the integral \( \int_{0}^{1} \frac{\tan^{-1} x}{1+x^{2}} \ dx \). This integral involves an inverse trigonometric function, specifically \( \tan^{-1} x \), and its derivative. To solve this, we will identify a suitable substitution.

Step 2: Choose a Suitable Substitution

Notice that the derivative of \( \tan^{-1} x \) with respect to \( x \) is \( \frac{1}{1+x^2} \). Therefore, let us set \( u = \tan^{-1} x \). Consequently, \( du = \frac{1}{1+x^{2}} \, dx \). This simplifies the integral substantially.

Step 3: Change Limits of Integration

With the substitution \( u = \tan^{-1} x \), we must change the limits of integration. When \( x = 0 \), \( u = \tan^{-1} 0 = 0 \). When \( x = 1 \), \( u = \tan^{-1} 1 = \frac{\pi}{4} \). So, the integral becomes \( \int_{0}^{\pi/4} u \, du \).

Step 4: Integrate the New Integral

Now, integrate \( \int_{0}^{\pi/4} u \, du \). The antiderivative of \( u \) is \( \frac{u^2}{2} \). Thus, evaluate from 0 to \( \pi/4 \):\[ \frac{u^2}{2} \bigg|_0^{\pi/4} = \frac{(\pi/4)^2}{2} - \frac{0^2}{2} = \frac{\pi^2}{32}. \]

Step 5: State the Final Result

We have evaluated the integral successfully. The final result of \( \int_{0}^{1} \frac{\tan^{-1} x}{1+x^{2}} \, dx \) is \( \frac{\pi^2}{32} \).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions, like \( \tan^{-1} x \), are essential in calculus as they help solve integrals involving trigonometric expressions. These functions are the reverse operations of standard trigonometric functions, such as tangent, sine, and cosine. They return the angle whose trigonometric value is given. For example, \( \tan^{-1} x \) returns the angle whose tangent is \( x \).
In the context of integral calculus, inverse trigonometric functions often appear in integrals that involve angles and arcs. Knowing their derivatives and integrals can help simplify complex expressions. For instance, the derivative of \( \tan^{-1} x \) is \( \frac{1}{1+x^2} \), which is useful in integration by substitution. Understanding these derivatives is vital for manipulating integrals in calculus.

Substitution Method

The substitution method is a powerful technique in calculus used to simplify the process of integration. It involves replacing a variable in an integral with another variable, making the integral easier to evaluate.
In this exercise, the substitution method is employed by letting \( u = \tan^{-1} x \). This smart choice makes the integral more manageable because \( du = \frac{1}{1+x^{2}} \, dx \). This substitution simplifies the integral and transforms it into a basic form that is easier to compute.

• Identify a part of the integral that can be replaced.
• Substitute this part with a new variable.
• Change the limits of integration if it's a definite integral.

The substitution method is not only practical but also a fundamental concept in solving many calculus problems efficiently.

Definite Integrals

Definite integrals compute the accumulation of quantities, like area under a curve, over a specific interval. They are denoted by the integral sign \( \int \) with bounds, here from \( 0 \) to \( 1 \), indicating the interval.
Changing the limits of integration is crucial when using substitutions, as it ensures that the new variable, in this context \( u \), accurately describes the same interval. The substitution \( u = \tan^{-1} x \) changes the limits from \( x = 0 \) to \( x = 1 \) and the corresponding \( u \)-values from \( 0 \) to \( \frac{\pi}{4} \).

• Recognize the need to transform the bounds alongside the integrand.
• Evaluate the new integral using these transformed bounds.

The result is the net area, in the case of this problem, accurately captured as \( \int_{0}^{\pi/4} u \, du \). This correct transformation is fundamental to getting the final answer right.

Calculus Problems

Calculus problems, like this exercise, often require a deeper understanding and application of various techniques to evaluate integrals successfully.
They involve recognizing patterns, such as spotting inverse trigonometric functions, and deciding on appropriate strategies like substitution. These problems are designed to test one’s ability to combine different calculus concepts creatively. To solve them:

• Identify the type of integral and functions involved.
• Choose suitable mathematical methods (e.g., substitution).
• Change variables and integrate with new limits if necessary.

The goal of these exercises is not only to find numerical answers but also to enhance problem-solving skills by connecting ideas across different calculus concepts. Through practice, students become adept at navigating complex calculus problems with confidence.