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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 8, Problema 8.4.16

Use any method to evaluate the integrals in Exercises 15–38. Most will require trigonometric substitutions, but some can be evaluated by other methods.
∫ dx / (1 + x²)

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1
Recognize that the integral \( \int \frac{dx}{1 + x^2} \) is a standard form that corresponds to the derivative of the inverse tangent function.
Recall the formula: \( \frac{d}{dx} \left( \arctan x \right) = \frac{1}{1 + x^2} \). This means the integral of \( \frac{1}{1 + x^2} \) with respect to \( x \) is \( \arctan x + C \), where \( C \) is the constant of integration.
Set up the integral as \( \int \frac{dx}{1 + x^2} = \arctan x + C \).
No trigonometric substitution is necessary here because the integrand matches the derivative of \( \arctan x \) directly.
Write the final answer as \( \arctan x + C \), where \( C \) represents the constant of integration.

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Trigonometric Substitution

Trigonometric substitution is a technique used to simplify integrals involving expressions like √(a² - x²), √(a² + x²), or √(x² - a²) by substituting x with a trigonometric function. This transforms the integral into a trigonometric integral, which is often easier to evaluate.
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Introduction to Trigonometric Functions

Integral of 1/(1 + x²)

The integral of 1/(1 + x²) with respect to x is a standard form that equals arctangent of x plus a constant. Recognizing this form allows direct evaluation without substitution, as ∫ dx/(1 + x²) = arctan(x) + C.
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Inverse Trigonometric Functions

Inverse trigonometric functions, such as arctan(x), arise naturally when integrating rational functions involving quadratic expressions. Understanding their derivatives and integrals helps in solving integrals that result in these functions.
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