Problem 19 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(1-34\) without using tables. $$ \int_{0}^{\infty} \frac{d v}{\left(1+v^{2}\right)\left(1+\tan ^{-1} v\right)} $$

Step-by-Step Solution

Answer

\( \ln\left(1 + \frac{\pi}{2}\right) \)

Step 1: Recognize the Form of the Integral

We have the integral \( \int_{0}^{\infty} \frac{d v}{(1+v^{2})(1+\tan^{-1}v)} \). The function \( \tan^{-1}v \) is the inverse tangent function, and \( 1+v^2 \) suggests similarity to the derivative of \( \tan^{-1}v \), which is \( \frac{1}{1+v^2} \).

Step 2: Substitute to Simplify

Notice that the derivative of \( \tan^{-1}v \) is \( \frac{1}{1+v^2} \). To simplify integration, consider the substitution \( u = \tan^{-1}v \). Then \( du = \frac{1}{1+v^2} dv \).

Step 3: Change the Limits of Integration

When \( v = 0 \), \( u = \tan^{-1}(0) = 0 \), and as \( v \to \infty \), \( u \to \frac{\pi}{2} \). Thus, the limits of integration change from \( v = 0 \) and \( v = \infty \) to \( u = 0 \) and \( u = \frac{\pi}{2} \).

Step 4: Rewrite the Integral

Using the substitution \( u = \tan^{-1}v \), the integral becomes:\[\int_{0}^{\frac{\pi}{2}} \frac{1}{1+u} \, du\]

Step 5: Integrate with Respect to the New Variable

The integral \( \int \frac{1}{1+u} \, du \) is a basic logarithmic integral. Its antiderivative is \( \ln|1+u| + C \). Evaluating this from \( 0 \) to \( \frac{\pi}{2} \), we have:\[\left[ \ln(1+u) \right]_{0}^{\frac{\pi}{2}} = \ln\left(1 + \frac{\pi}{2}\right) - \ln(1) = \ln\left(1 + \frac{\pi}{2}\right)\]

Step 6: Simplify the Result

The expression simplifies to \( \ln\left(1 + \frac{\pi}{2}\right) \). This is the final result of the original integral.

Key Concepts

Substitution Method

The Substitution Method is a smart technique used in calculus to simplify complex integrals. It involves changing variables to make the integral easier to solve. In essence, you're replacing a complicated part of the function with a simpler variable.
For example, suppose you're working with an integral that includes an inverse trigonometric function, like in our exercise. You can often use substitution if you notice that one part of the integral's derivative matches another part of the function.

• Identify the part of the integral that can be substituted. Usually, this won't just be any term, but rather the one whose derivative appears somewhere else in the integral.
• Rewrite the integral in terms of the new variable using the derivative.
• Don't forget to adjust the limits of integration if you're dealing with definite integrals.

In our example, we made the substitution \( u = \tan^{-1} v \), which transformed the difficult integral into a simpler one.

Definite Integrals

Definite Integrals calculate the area under a curve within specific limits, and they're a key part of calculus.
This idea helps us not just find areas, but also solve real-world problems involving cumulative quantities, like distance and total cost.
When you encounter a definite integral, you'll usually see it represented like this: \[ \int_{a}^{b} f(x) \, dx\]This tells us to integrate the function \( f(x) \) as \( x \) goes from \( a \) to \( b \). The limits \( a \) and \( b \) are where the integration starts and ends.

• One must evaluate the antiderivative of the function over the range \( a \) and \( b \).
• Subtract the value of the antiderivative at the lower limit from the value of the antiderivative at the upper limit.

In our specific example, the limits changed from \( 0 \) to \( \infty \) as \( v \) changes, to \( 0 \) and \( \frac{\pi}{2} \) in the "u" world after substitution.

Inverse Trigonometric Functions

Inverse Trigonometric Functions, like \( \tan^{-1}v \), are functions that reverse the trigonometric ratios. They are helpful when you need to find an angle when given a ratio.
Each trigonometric function has its inverse:

• \( \sin^{-1}(x) \) or arcsin(x)
• \( \cos^{-1}(x) \) or arccos(x)
• \( \tan^{-1}(x) \) or arctan(x)

These functions can be tricky to integrate directly. Often, their derivatives appear in different parts of an integral, which is the key to using substitution.
Take \( \tan^{-1}v \): its derivative \( \frac{1}{1+v^2} \) was crucial in the problem, allowing us to perform our substitution effectively.
Keep in mind:

• Recognize patterns connecting trigonometric functions and their derivatives.
• Remember that these functions help express the relationships between angles and ratios.

Understanding these can simplify many calculus problems and help transition between geometric and algebraic interpretations of a problem.