int_(2)^(4)(2)/(xsqrt(x^(2)-4))dx

Substitution and Simplification: Substitute x = 2sec(θ) into the integral and simplify. ∫(2)/(x√(x^2-4))dx = ∫(2)/(2sec(θ)√(4sec^2(θ)-4)) * 2sec(θ)tan(θ)dθ = ∫(2)/(2sec(θ)√(4(tan^2(θ)+1)-4)) * 2sec(θ)tan(θ)dθ = ∫(2)/(2sec(θ)√(4tan^2(θ))) * 2sec(θ)tan(θ)dθ = ∫(2)/(2sec(θ) * 2tan(θ)) * 2sec(θ)tan(θ)dθ = ∫(2)/(4tan(θ)) * 2sec(θ)tan(θ)dθ = ∫(1/2)dθIntegration and Simplification: Integrate (1/2)dθ from 0 to π/3. ∫(1/2)dθ = (1/2)θ Evaluating from 0 to π/3 gives us: (1/2)θ | from 0 to π/3 = (1/2)(π/3) - (1/2)(0) = π/6Evaluation and Conversion: Convert the result back to x. Since we made the substitution x = 2sec(θ), we need to find the corresponding θ values for x = 2 and x = 4. However, we already have these values from the substitution step: θ = 0 when x = 2 and θ = π/3 when x = 4. Therefore, the definite integral from 2 to 4 of the function (2)/(x√(x^2-4)) with respect to x is π/6.

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$\int_{2}^{4}\frac{2}{x\sqrt{x^{2}-4}}dx$

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Calculus

Find indefinite integrals using the substitution and by parts

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\int_{2}^{4}\frac{2}{x\sqrt{x^{2}-4}}dx

Substitution and Simplification: Substitute $x = 2\sec(\theta)$ into the integral and simplify. $\int\frac{2}{x\sqrt{x^2-4}}dx = \int\frac{2}{2\sec(\theta)\sqrt{4\sec^2(\theta)-4}} \cdot 2\sec(\theta)\tan(\theta)d\theta$ $= \int\frac{2}{2\sec(\theta)\sqrt{4(\tan^2(\theta)+1)-4}} \cdot 2\sec(\theta)\tan(\theta)d\theta$ $= \int\frac{2}{2\sec(\theta)\sqrt{4\tan^2(\theta)}} \cdot 2\sec(\theta)\tan(\theta)d\theta$ $= \int\frac{2}{2\sec(\theta) \cdot 2\tan(\theta)} \cdot 2\sec(\theta)\tan(\theta)d\theta$ $= \int\frac{2}{4\tan(\theta)} \cdot 2\sec(\theta)\tan(\theta)d\theta$ $= \int\left(\frac{1}{2}\right)d\theta$
Integration and Simplification: Integrate $(\frac{1}{2})d\theta$ from $0$ to $\frac{\pi}{3}$ .
$\int(\frac{1}{2})d\theta = (\frac{1}{2})\theta$
Evaluating from $0$ to $\frac{\pi}{3}$ gives us:
$(\frac{1}{2})\theta |$ from $0$ to $\frac{\pi}{3}$ = $(\frac{1}{2})(\frac{\pi}{3}) - (\frac{1}{2})(0) = \frac{\pi}{6}$
Evaluation and Conversion: Convert the result back to $x$ . Since we made the substitution $x = 2\sec(\theta)$ , we need to find the corresponding $\theta$ values for $x = 2$ and $x = 4$ . However, we already have these values from the substitution step: $\theta = 0$ when $x = 2$ and $\theta = \frac{\pi}{3}$ when $x = 4$ . Therefore, the definite integral from $2$ to $x = 2\sec(\theta)$ $0$ of the function $x = 2\sec(\theta)$ $1$ with respect to $x$ is $x = 2\sec(\theta)$ $3$ .