Question

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

$$\int \frac{2}{x \sqrt{4 x^{2}-1}} d x$$ The result of the given indefinite integral is: $$\ln|x| + C$$

Step-by-step solution

Identify the substitution

Let \(u = 2x\). Then, we get: $$\frac{du}{2} = dx$$ This substitution will allow us to simplify the terms under the square root in the denominator. 2. Perform the substitution

Perform the substitution

Making the substitution we get: $$\int \frac{2}{x \sqrt{4 x^{2}-1}} d x = \int \frac{1}{x \sqrt{u^{2}-1}} du$$ 3. Rewrite the integral using trigonometric substitution

Rewrite the integral using trigonometric substitution

Let \(u = \cosh(t)\), then we have \(du = \sinh(t) dt\). The integral becomes: $$\int \frac{1}{x \sqrt{u^{2}-1}} du = \int \frac{1}{x \sinh(t)} \sinh(t) dt = \int \frac{1}{x} dt$$ 4. Evaluate the integral

Evaluate the integral

Now we can evaluate the integral: $$\int \frac{1}{x} dt = \ln|x| + C$$ 5. Substitute back for \(u\) and \(x\)

Substitute back for \(u\) and \(x\)

Now, substitute back for \(u\) and \(x\). We have: $$\ln|x| + C = \ln\left|\frac{1}{2}\cosh(t)\right| + C$$ 6. Use the inverse hyperbolic function to find \(t\)

Use the inverse hyperbolic function to find \(t\)

Recall that \(u = \cosh(t)\). Then, \(t = \cosh^{-1}(u)\), and we can write: $$\ln\left|\frac{1}{2}\cosh(t)\right| + C = \ln\left|\frac{1}{2}\cosh(\cosh^{-1}(u))\right| + C = \ln\left|\frac{1}{2}u\right| + C$$ 7. Substitute back for \(x\)

Substitute back for \(x\)

Recall that \(u = 2x\). Then, we can write: $$\ln\left|\frac{1}{2}u\right| + C = \ln\left|\frac{1}{2}(2x)\right| + C = \ln|x| + C$$ 8. Check the result by differentiation

Check the result by differentiation

Take the derivative of our final result with respect to \(x\) to verify that it matches the original integrand: $$\frac{d}{dx}(\ln|x| + C) = \frac{1}{x}$$ Since the derivative matches the original integrand, our result is correct. The indefinite integral evaluates to: $$\int \frac{2}{x \sqrt{4 x^{2}-1}} d x = \ln|x| + C$$

Key concepts

Substitution Method

The substitution method is a powerhouse in calculus, especially when encountering integrals that are not straightforward. Think of it like a strategy game where we use a tactical maneuver, swapping the challenging parts of the integral with simpler expressions to allow an easier integration path forward.

Here's how it works: identify a part of the integral that can be replaced with a variable, generally referred to as 'u'. This move simplifies the integral into one that's more manageable. For example, if you spot a function inside another function, that's your cue for a potential 'u' substitution! With the function identified and 'u' assigned, we rewrite the entire integral in terms of 'u'.

The next step is to find the derivative of 'u', designated as 'du', and replace the differential part of the integral (like 'dx' or 'dt') with this new 'du'. Once the substitution is made, the integral is often much easier to solve. After integrating with respect to 'u', don't forget the final, crucial step: substitute back in the original variables to complete the solution.

Trigonometric Substitution

Trigonometric substitution is a clever trick used when an integral contains a square root of a quadratic expression. It's like solving a puzzle by fitting trigonometric functions into the integral to simplify it.

Typically, the choice of trigonometric substitution depends on the form of the quadratic under the square root. For example, \(\sqrt{a^2 - x^2}\) suggests a sine substitution, \(\sqrt{a^2 + x^2}\) a tangent substitution, and \(\sqrt{x^2 - a^2}\) a secant or hyperbolic cosine substitution. By replacing the troublesome part with a trigonometric function, you transform the integral into a friendlier form that involves familiar trigonometric identities.

After the substitution and solving the integral in trigonometric terms, you must revert back to the original variable by using the inverse trigonometric functions. And voilà, you have tamed a once-difficult problem into a solvable one. It's like finding the right key for a locked treasure chest.

Hyperbolic Functions

Hyperbolic functions are the unsung heroes of integration, often resembling trigonometric functions but with a twist. They include functions like \(\sinh(x)\), \(\cosh(x)\), and \(\tanh(x)\), which arise naturally in many contexts within calculus.

These functions bear some striking similarities to the trigonometric functions but differ in important ways. For instance, their definitions involve the exponential function \(e^x\), and they represent hyperbolas, rather than circles or ellipses. They are especially useful for integrals involving square roots of the form \(\sqrt{x^2 + 1}\) or \(\sqrt{x^2 - 1}\).

Using hyperbolic functions for substitution can streamline an integral that seems insurmountable. After the substitution, identifying the right hyperbolic identity or derivative allows the integration to proceed smoothly. And just as with trigonometric functions, we have inverse hyperbolic functions that help us to reverse the substitution, unveiling the integral's result within the realm of algebraic expressions.