SOLUTION: What is the solution to the integral of (ln^3 x) / (x)(√ln^2 x - 4) dx using trigonometric substitution? The answer should be equal to ((8+ln^2 x)/3)(√ln^2 x - 4).
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Question 1193654: What is the solution to the integral of (ln^3 x) / (x)(√ln^2 x - 4) dx using trigonometric substitution? The answer should be equal to ((8+ln^2 x)/3)(√ln^2 x - 4).
Answer by parmen(42) (Show Source): You can put this solution on YOUR website!
Certainly, let's tackle this integral using trigonometric substitution.
**1. Preparation**
* **Substitution:**
* Let ln(x) = 2sec(θ)
* Then, d(ln(x)) = 2sec(θ)tan(θ) dθ
* Which implies: (1/x) dx = 2sec(θ)tan(θ) dθ
**2. Substitute into the Integral**
* The integral becomes:
∫ (ln³(x)) / (x)(√(ln²(x) - 4)) dx
= ∫ (2sec(θ))³ * 2sec(θ)tan(θ) / (√(4sec²(θ) - 4)) dθ
= ∫ 16sec⁴(θ)tan(θ) / (2√(sec²(θ) - 1)) dθ
= ∫ 8sec⁴(θ)tan(θ) / tan(θ) dθ
= ∫ 8sec⁴(θ) dθ
**3. Simplify the Integral**
* Use the identity: sec²(θ) = 1 + tan²(θ)
∫ 8sec⁴(θ) dθ = ∫ 8(1 + tan²(θ))sec²(θ) dθ
= ∫ 8sec²(θ) dθ + ∫ 8tan²(θ)sec²(θ) dθ
**4. Evaluate the Integrals**
* ∫ 8sec²(θ) dθ = 8tan(θ) + C₁
* ∫ 8tan²(θ)sec²(θ) dθ = 8/3 * tan³(θ) + C₂
**5. Combine the Results**
* ∫ 8sec⁴(θ) dθ = 8tan(θ) + 8/3 * tan³(θ) + C
**6. Back-Substitute for x**
* Recall: ln(x) = 2sec(θ)
=> sec(θ) = ln(x) / 2
=> tan(θ) = √(sec²(θ) - 1) = √((ln²(x) / 4) - 1) = √(ln²(x) - 4) / 2
* Substitute these values back into the result:
∫ (ln³(x)) / (x)(√(ln²(x) - 4)) dx = 8 * (√(ln²(x) - 4) / 2) + 8/3 * (√(ln²(x) - 4) / 2)³ + C
= 4√(ln²(x) - 4) + (1/3) * √(ln²(x) - 4)³ + C
**Therefore, the solution to the integral is:**
∫ (ln³(x)) / (x)(√(ln²(x) - 4)) dx = 4√(ln²(x) - 4) + (1/3) * √(ln²(x) - 4)³ + C
I hope this comprehensive solution is helpful!
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