Question

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

Short answer

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

Step-by-step solution

Identify the substitution

From the function $\frac{2}{x\sqrt{4x^2-1}}$, 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 $du$.

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 $dx$: $$dx=\frac{1}{2}\sinh udu$$

Substitute x and dx in the integral

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

Simplify the integral

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

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: $$\int \frac{1}{\cosh u}du={\tanh }^{-1}(\tanh u)+C$$

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: $$\tanh u=\frac{\sinh u}{x}$$ $${\tanh }^{-1}(\tanh u)+C={\tanh }^{-1}\left(\frac{2\sinh u}{\cosh u}\right)+C={\tanh }^{-1}\left(\frac{2\sqrt{x^2-\frac{1}{4}}}{x}\right)+C$$

Check the result by differentiating

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