Problem 20 Converge. Evaluate the integrals... [FREE SOLUTION]

Chapter 8: Problem 20

Converge. Evaluate the integrals without using tables. $$\int_{0}^{\infty} \frac{16 \tan ^{-1} x}{1+x^{2}} d x$$

Short Answer

Expert verified

The integral evaluates to $ 2\pi^2 $.

Step by step solution

Set up the Integral

We begin with the integral $ \int_{0}^{\infty} \frac{16 \tan^{-1} x}{1+x^{2}} \, dx $. Our goal is to find the value of this improper integral.

Change of Variable

We will use the substitution $ u = \tan^{-1}(x) $, which means that $ x = \tan(u) $ and $ dx = \sec^2(u) \, du $. The limits of integration change accordingly: when $ x = 0 $, $ u = \tan^{-1}(0) = 0 $, and when $ x \to \infty $, $ u = \tan^{-1}(\infty) = \frac{\pi}{2} $.

Substitute in the Integral

Substituting $ u = \tan^{-1}(x) $ into the integral, we get:\[\int_{0}^{\pi/2} 16u \cdot \frac{1}{1+\tan^2(u)} \cdot \sec^2(u) \, du\]Note that $ \frac{1}{1+\tan^2(u)} = \cos^2(u) $, so the expression simplifies to:\[\int_{0}^{\pi/2} 16u \, du\]

Evaluate the Integral

The integral $ \int 16u \, du $ from 0 to $ \pi/2 $ becomes a standard definite integral:\[\left[ 8u^2 \right]_{0}^{\pi/2} = 8 \left( \frac{\pi}{2} \right)^2 - 8(0)^2 = 8 \cdot \frac{\pi^2}{4}\]Simplifying gives us $ 2\pi^2 $.

Conclusion

The evaluated improper integral is $ 2\pi^2 $. This shows that the integral converges to this value.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integral Evaluation

Evaluating an integral involves finding the total area under the curve of a function within a given interval. When dealing with improper integrals like $ \int_{0}^{\infty} \frac{16 \tan^{-1} x}{1+x^{2}} \, dx $, the limits extend to infinity, requiring careful evaluation. Such integrals can converge to a finite value or diverge.

Start by setting up the integral to identify its components.
Determine if the integral needs any substitutions or transformations for simplification.
After simplifying, perform the integration calculations.
Finally, interpret the result to conclude whether the integral converges or diverges.

Substitution Method

The substitution method is a technique used to simplify integrals, making them easier to evaluate. In this exercise, we used the substitution $ u = \tan^{-1}(x) $. This turns the integral $ \int_{0}^{\infty} \frac{16 \tan^{-1} x}{1+x^{2}} \, dx $ into an easier form by changing variables.

Substituting $ u = \tan^{-1}(x) $ means $ x = \tan(u) $, requiring us to express $ dx $ in terms of $ du $ as $ dx = \sec^2(u) \, du $.
The limits of integration change from $ x = 0 $ and $ x = \infty $ to $ u = 0 $ and $ u = \frac{\pi}{2} $.
This results in a simpler integral expression that is easier to handle.

Substitution is often used to transform the integral into a more manageable form for evaluation.

Arc Tangent Function

The arc tangent function, denoted as $ \tan^{-1}(x) $, is the inverse of the tangent function. This function plays a central role in this integral, influencing both substitution and simplification.

The derivative of $ \tan^{-1}(x) $ is $ \frac{1}{1+x^2} $, which appears in the integrand.
Changing the variable to $ u = \tan^{-1}(x) $ helps utilize its properties.

The properties of the arc tangent are crucial for dealing with integrals that feature them prominently.

Limits of Integration

The limits of integration define the interval over which an integral is evaluated. For improper integrals like $ \int_{0}^{\infty} $, it involves a careful approach as the upper limit extends to infinity.

We change from $ x $ to $ u = \tan^{-1}(x) $, modifying the limits to $ 0 $ and $ \frac{\pi}{2} $ after substitution.
Adjusting limits correctly is vital to ensure accurate computation.
Make sure to check whether the limits lead to a convergence or divergence.

Proper handling of limits is necessary to ensure that integrals are evaluated effectively and correctly.

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91影视

Converge. Evaluate the integrals without using tables. $$\int_{0}^{\infty} \frac{16 \tan ^{-1} x}{1+x^{2}} d x$$

Short Answer

Step by step solution

Set up the Integral

Change of Variable

Substitute in the Integral

Evaluate the Integral

Conclusion

Key Concepts

Integral Evaluation

Substitution Method

Arc Tangent Function

Limits of Integration

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