7-56. Trigonometric substitutions Evaluate the following integrals using trigonometric substitution.
19. ∫ 1/√(x² - 81) dx, x > 9

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Recognize that the integral involves a square root of the form √(x² - a²), which suggests using the trigonometric substitution x = a sec(θ). Here, a = 9, so substitute x = 9 sec(θ).

Differentiate x = 9 sec(θ) to find dx. Using the derivative of sec(θ), dx = 9 sec(θ) tan(θ) dθ.

Substitute x = 9 sec(θ) and dx = 9 sec(θ) tan(θ) dθ into the integral. Also, replace √(x² - 81) using the identity sec²(θ) - 1 = tan²(θ), which simplifies √(x² - 81) to 9 tan(θ).

Simplify the integral after substitution. The square root √(x² - 81) becomes 9 tan(θ), and dx becomes 9 sec(θ) tan(θ) dθ. The integral simplifies to ∫ (1 / (9 tan(θ))) * (9 sec(θ) tan(θ)) dθ.

Evaluate the simplified integral, which reduces to ∫ sec(θ) dθ. The antiderivative of sec(θ) is ln|sec(θ) + tan(θ)| + C. Finally, revert back to the original variable x using the substitution x = 9 sec(θ) and the trigonometric relationships.

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