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

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Full solution

Q. ∫3∞1xx2−9dx \int_{3}^{\infty}\frac{1}{x\sqrt{x^{2}-9}}dx
  1. Recognize standard form: Recognize the integral as a standard form. The integral given is ∫3∞1xx2−9dx\int_{3}^{\infty} \frac{1}{x\sqrt{x^2-9}}dx. This is a standard form of an improper integral that can be solved using a trigonometric substitution.
  2. Perform trigonometric substitution: Perform a trigonometric substitution. Let x=3sec⁡(θ)x = 3\sec(\theta), where θ\theta is from 00 to π2\frac{\pi}{2} as xx goes from 33 to infinity. Then, dx=3sec⁡(θ)tan⁡(θ)dθdx = 3\sec(\theta)\tan(\theta)d\theta and x2−9=9sec⁡2(θ)−9=3tan⁡(θ)\sqrt{x^2-9} = \sqrt{9\sec^2(\theta)-9} = 3\tan(\theta).
  3. Substitute xx and dxdx: Substitute xx and dxdx in the integral.\newlineThe integral becomes ∫13sec⁡(θ)⋅3tan⁡(θ)⋅3sec⁡(θ)tan⁡(θ) dθ\int \frac{1}{3\sec(\theta)\cdot 3\tan(\theta)} \cdot 3\sec(\theta)\tan(\theta)\,d\theta from 00 to π2\frac{\pi}{2}. The 3sec⁡(θ)tan⁡(θ)3\sec(\theta)\tan(\theta) terms cancel out, simplifying the integral to ∫13 dθ\int \frac{1}{3}\,d\theta from 00 to π2\frac{\pi}{2}.
  4. Integrate with respect: Integrate with respect to θ\theta. The integral is now (1/3)∫dθ(1/3)\int d\theta from 00 to π/2\pi/2, which is simply (1/3)θ(1/3)\theta evaluated from 00 to π/2\pi/2.
  5. Evaluate integral: Evaluate the integral. Plugging in the limits of integration, we get (13)(π2−0)=π6(\frac{1}{3})(\frac{\pi}{2} - 0) = \frac{\pi}{6}.
  6. Write final answer: Write the final answer.\newlineThe value of the integral from 33 to infinity of 1xx2−9\frac{1}{x\sqrt{x^2-9}}dx is π6\frac{\pi}{6}.

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