{Use of Tech} Using the integral of sec³u By reduction formula 4 in Section 8.3,
∫sec³u du = 1/2 (sec u tan u + ln |sec u + tan u|) + C

Graph the following functions and find the area under the curve on the given interval.
f(x) = 1/(x√(x² - 36)), [12/√3 , 12]

검증된 단계별 안내

First, rewrite the integral for the area under the curve as \(\int_{12/\sqrt{3}}^{12} \frac{1}{x \sqrt{x^2 - 36}} \, dx\).

Use the substitution \(x = 6 \sec u\) because \(\sqrt{x^2 - 36}\) suggests a secant substitution (since \(36 = 6^2\)). Then, compute \(dx\) in terms of \(du\).

Express the integral entirely in terms of \(u\) by substituting \(x = 6 \sec u\), \(dx = 6 \sec u \tan u \, du\), and \(\sqrt{x^2 - 36} = 6 \tan u\).

Simplify the integral after substitution to get an integral involving \(\sec^3 u\), which matches the form given in the reduction formula: \(\int \sec^3 u \, du\).

Apply the reduction formula \(\int \sec^3 u \, du = \frac{1}{2} (\sec u \tan u + \ln |\sec u + \tan u|) + C\) to evaluate the integral, then substitute back to \(x\) using \(u = \sec^{-1}(x/6)\) and evaluate the definite integral at the given limits.

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주요 개념

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Integration Using Reduction Formulas

Reduction formulas simplify complex integrals by expressing them in terms of simpler integrals of the same type. For example, the integral of sec³u can be evaluated using a known reduction formula, which breaks it down into a combination of sec u tan u and a logarithmic term. This technique is essential for handling powers of trigonometric functions.

Definite Integrals and Area Under a Curve

A definite integral calculates the net area between a function's graph and the x-axis over a specified interval. Evaluating ∫ f(x) dx from a to b gives the total area under f(x) between x = a and x = b, accounting for regions above and below the axis. This concept connects integration to geometric interpretation.

Trigonometric Substitution in Integration

Trigonometric substitution is a method used to simplify integrals involving expressions like √(x² - a²) by substituting x with a trigonometric function (e.g., x = a sec θ). This transforms the integral into a trigonometric integral, which can be easier to evaluate using known identities and formulas, such as the integral of sec³u.

{Use of Tech} Using the integral of sec³u By reduction formula 4 in Section 8.3,∫sec³u du = 1/2 (sec u tan u + ln |sec u + tan u|) + CGraph the following functions and find the area under the curve on the given interval.f(x) = 1/(x√(x² - 36)), [12/√3 , 12]