int_(3)^(oo)(1)/(xsqrt(x^(2)-9))dx

Question

Bytelearn · Accepted Answer

Recognize standard form: Recognize the integral as a standard form. The integral given is ∫(1)/(x*sqrt(x^2-9))dx from 3 to infinity. This is a standard form of an improper integral that can be solved using a trigonometric substitution.Perform trigonometric substitution: Perform a trigonometric substitution. Let x = 3sec(θ), where θ is from 0 to π/2 as x goes from 3 to infinity. Then, dx = 3sec(θ)tan(θ)dθ and sqrt(x^2-9) = sqrt(9sec^2(θ)-9) = 3tan(θ).Substitute x and dx: Substitute x and dx in the integral. The integral becomes ∫(1)/(3sec(θ)*3tan(θ)) * 3sec(θ)tan(θ)dθ from 0 to π/2. The 3sec(θ)tan(θ) terms cancel out, simplifying the integral to ∫(1)/(3)dθ from 0 to π/2.Integrate with respect: Integrate with respect to θ. The integral is now (1/3)∫dθ from 0 to π/2, which is simply (1/3)θ evaluated from 0 to π/2.Evaluate integral: Evaluate the integral. Plugging in the limits of integration, we get (1/3)(π/2 - 0) = π/6.Write final answer: Write the final answer. The value of the integral from 3 to infinity of (1)/(x*sqrt(x^2-9))dx is π/6.

$\int_{3}^{\infty}\frac{1}{x\sqrt{x^{2}-9}}dx$

Full solution

More problems from Find indefinite integrals using the substitution and by parts

Related topics

∫3∞1xx2−9dx \int_{3}^{\infty}\frac{1}{x\sqrt{x^{2}-9}}dx ∫3∞​xx2−9​1​dx

$\int_{3}^{\infty}\frac{1}{x\sqrt{x^{2}-9}}dx$