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Chapter 5: Problem 92

Evaluate the following definite integrals using the Fundamental Theorem of Calculus. $$\int_{\sqrt{2}}^{2} \frac{d x}{x \sqrt{x^{2}-1}}$$

Short Answer

Expert verified
Question: Evaluate the definite integral \(\int_{\sqrt{2}}^{2} \frac{dx}{x \sqrt{x^{2}-1}}\). Answer: \(\frac{1}{2}\left[\arcsin(3) - \arcsin(1)\right]\)

Step by step solution

01

Using substitution

We will first find the antiderivative using substitution. Let \(u = x^2 - 1\), then we get, $$ \frac{du}{dx} = 2x \Rightarrow dx =\frac{du}{2x} $$ Now, rewrite the integral in terms of \(u\), $$\int_{\sqrt{2}}^{2} \frac{dx}{x \sqrt{x^{2}-1}} = \int_{1}^{3}\frac{du}{2x \sqrt{u}}$$ Note that we have also changed the limits of integration to match \(u\).
02

Simplify the integral in terms of u

Now observe that \(x = \sqrt{u+1}\), then rewrite the integral as, $$\int_{1}^{3}\frac{du}{2 \sqrt{u+1}\sqrt{u}} = \frac{1}{2} \int_{1}^{3} \frac{du}{\sqrt{u^{2}+u}}$$
03

Find the antiderivative

Now integrate the simplified integral. Use the fact that the antiderivative of \(\frac{1}{\sqrt{1+u^2}}\) is \(\arcsin{u}\), we get $$\frac{ant}{2}[\arcsin(u)]_{1}^{3}$$
04

Evaluate the limits

Finally, evaluate the definite integral by substituting the limits, $$\frac{1}{2}\left[\arcsin(3) - \arcsin(1)\right]$$ So, the value of the given definite integral is: $$ \int_{\sqrt{2}}^{2} \frac{d x}{x \sqrt{x^{2}-1}} = \frac{1}{2}\left[\arcsin(3) - \arcsin(1)\right]$$

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