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Chapter 5: Problem 32

Indefinite integrals Use a change of variables or Table 5.6 to evaluate the following indefinite integrals. Check your work by differentiating. $$\int \frac{2}{x \sqrt{4 x^{2}-1}} d x, x>\frac{1}{2}$$

Short Answer

Expert verified
Question: Evaluate the indefinite integral $$\int \frac{2}{x \sqrt{4 x^{2}-1}} dx$$. Answer: The indefinite integral evaluates to $$\tanh^{-1}\left(\frac{2\sqrt{x^2-\frac{1}{4}}}{x}\right)+C$$.

Step by step solution

01

Identify the substitution

From the function \(\frac{2}{x \sqrt{4 x^{2}-1}}\), we see it is best to use substitution with trigonometric functions as the denominator resembles the Pythagorean identity. Let's use the substitution: $$x=\frac{1} {2} \cosh u$$ Now, we will find the differential \(dx\) in terms of \(du\).
02

Find dx in terms of du

Differentiate x with respect to u, $$\frac{dx}{du}=\frac{1}{2}\sinh u$$ Then, multiply both sides by \(du\) to find \(dx\): $$dx=\frac{1}{2}\sinh u \ du$$
03

Substitute x and dx in the integral

Now we will substitute the expressions for x and dx in the integral: $$\int \frac{2}{x \sqrt{4 x^{2}-1}} d x = \int \frac{2}{\frac{1}{2} \cosh u \sqrt{4 (\frac{1}{2}\cosh u)^{2}-1}}\cdot (\frac{1}{2}\sinh u\ du)$$
04

Simplify the integral

Combine the terms in the integral and simplify: $$\int \frac{2(\frac{1}{2}\sinh u)}{\frac{1}{2} \cosh u \sqrt{1 - \frac{1}{4}\cosh^2 u}} du = \int \frac{\sinh u}{\cosh u \sqrt{1 - \frac{1}{4}\cosh^2 u}} du$$
05

Evaluate the integral

By observing the integral, we can write \(\cosh^2 u - 1 = \sinh^2 u\). So, we replace the denominator with the new expression: $$\int \frac{\sinh u}{\cosh u \sqrt{\sinh^2 u}} du = \int \frac{\sinh u}{\cosh u \cdot (\sinh u)} du$$ $$=\int \frac{1}{\cosh u}du$$ We recognize this integral as the inverse hyperbolic tangent function: $$\int \frac{1}{\cosh u}du= \tanh^{-1}(\tanh u) + C$$
06

Re-substitute x in terms of u

Now, recall the substitution we made: $$x=\frac{1}{2}\cosh u$$ So, we will use the definition of hyperbolic tangent function \(\tanh u = \frac{\sinh u}{\cosh u}\) and rewrite the result in terms of x: $$\tanh u = \frac{\sinh u}{x}$$ $$\tanh^{-1}(\tanh u) + C=\tanh^{-1}\left(\frac{2\sinh u}{\cosh u}\right) + C= \tanh^{-1}\left(\frac{2\sqrt{x^2-\frac{1}{4}}}{x}\right)+C$$
07

Check the result by differentiating

We will now differentiate the result with respect to x to confirm that we get the original integrand: $$\frac{d}{dx}\left(\tanh^{-1}\left(\frac{2\sqrt{x^2-\frac{1}{4}}}{x}\right)+C\right) = \frac{2}{x \sqrt{4 x^{2}-1}}$$ The derivative is equal to the original integrand function, confirming that our solution is correct.

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