Question

Evaluate the integrals in Exercises 31–78.67. ∫(from -2 to 2)3dt/(4+3t²)

Accepted Answer

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 $$f(-t) = f(t). $$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}$$} \, dt. $$Factor constants out of the integral where possible: $$2 	imes 3 = 6$$, so the integral becomes $$6 \int_{0}^{2} \frac{1}{4 + 3t$$^{2}$$} \, dt. To $$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}$$. Use the substitution u$$ = \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} ight) + C$$. Then substitute back and evaluate the definite integral from t=0$ to t=2$.

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

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Key Concepts

Definite Integrals

Integration of Rational Functions

Even Functions and Symmetry in Integration