Question

Evaluate the following integrals. Consider completing the square. $$\int_{2+\sqrt{2}}^{4} \frac{d x}{\sqrt{(x-1)(x-3)}}$$

Short answer

Question: Evaluate the definite integral: $$\int_{2+\sqrt{2}}^{4} \frac{d x}{\sqrt{(x-1)(x-3)}}$$ Answer: The value of the definite integral is: $$\int_{2+\sqrt{2}}^{4} \frac{d x}{\sqrt{(x-1)(x-3)}} = 2 - \sqrt{2}$$

Step-by-step solution

Complete the square of the denominator

First, let's rewrite the integrand by replacing the expression inside the square root with a perfect square: The expression to complete the square is \((x-1)(x-3) = x^2-4x+3\). We need to write this in the form \((x-a)^2 + b\). To find 'a', take half the coefficient of 'x', which is -4. Half of -4 is -2. Now we have \((x-2)^2 + b\). To find 'b', substitute 'a' back into the equation and subtract: \(b = 3 - (-2)^2 = 3 - 4 = -1\). So our expression becomes: $$\int_{2+\sqrt{2}}^{4} \frac{d x}{\sqrt{(x-2)^2 - 1}}$$

Perform a substitution

We'll use the substitution method to simplify the integrand. Let \(x = 2 + \sinh(u)\). Differentiate x with respect to u: \(\frac{dx}{du} = \cosh(u)\). Then, change the limits of integration accordingly: Lower limit: \(u =\sinh^{-1}(2+\sqrt{2} - 2)\), Upper limit: \(u =\sinh^{-1}(4 - 2)\). By substituting, we get: $$\int_{\sinh^{-1}(\sqrt{2})}^{\sinh^{-1}(2)} \frac{\cosh(u)}{\sqrt{\sinh^2(u) - 1}}du$$

Simplify the integrand and integrate

Our integrand simplifies to: $$\int_{\sinh^{-1}(\sqrt{2})}^{\sinh^{-1}(2)} \cosh(u)du$$ Now, we integrate with respect to u: $$\int \cosh(u)du = \sinh(u) + C$$

Apply the Fundamental Theorem of Calculus

To find the definite integral, we subtract the antiderivative evaluated at the lower limit from the antiderivative evaluated at the upper limit: $$\sinh(\sinh^{-1}(2)) - \sinh(\sinh^{-1}(\sqrt{2})) = 2 - \sqrt{2}$$ Hence, the value of the definite integral is: $$\int_{2+\sqrt{2}}^{4} \frac{d x}{\sqrt{(x-1)(x-3)}} = 2 - \sqrt{2}$$