Evaluate the integrals in Exercises 31–78.
67. ∫(from -2 to 2)3dt/(4+3t²)

Verified step by step guidance

1. Recognize that the integral is a definite integral of the form \(\int_{-2}^{2} \frac{3}{4 + 3t^{2}} \, dt\). Notice that the integrand is an even function because \(t^{2}\) is even, so \(f(-t) = f(t)\).
2. Use the property of even functions in definite integrals: \(\int_{-a}^{a} f(t) \, dt = 2 \int_{0}^{a} f(t) \, dt\). This allows us to rewrite the integral as \(2 \int_{0}^{2} \frac{3}{4 + 3t^{2}} \, dt\).
3. Factor constants out of the integral where possible: \(2 \times 3 = 6\), so the integral becomes \(6 \int_{0}^{2} \frac{1}{4 + 3t^{2}} \, dt\).
4. To integrate \(\int \frac{1}{4 + 3t^{2}} \, dt\), rewrite the denominator to match the form \(a^{2} + u^{2}\), which is suitable for the arctangent integration formula. Express \(4 + 3t^{2}\) as \(2^{2} + (\sqrt{3} t)^{2}\).
5. Use the substitution \(u = \sqrt{3} t\), so \(dt = \frac{du}{\sqrt{3}}\). Rewrite the integral in terms of \(u\) and apply the formula \(\int \frac{1}{a^{2} + u^{2}} \, du = \frac{1}{a} \arctan \left( \frac{u}{a} \right) + C\). Then substitute back and evaluate the definite integral from \(t=0\) to \(t=2\).

Key Concepts

Definite Integrals

A definite integral calculates the net area under a curve between two specific limits. It is represented as ∫ from a to b of f(t) dt, where a and b are the lower and upper bounds. The result is a number representing the accumulated value of the function over that interval.

Integration of Rational Functions

Rational functions are ratios of polynomials, and integrating them often involves recognizing standard forms or using substitution. For integrals like ∫ 1/(a + bt²) dt, the antiderivative typically involves inverse trigonometric functions such as arctangent.

Even Functions and Symmetry in Integration

An even function satisfies f(-t) = f(t), meaning its graph is symmetric about the y-axis. When integrating an even function over symmetric limits [-a, a], the integral equals twice the integral from 0 to a, simplifying calculations.