Problem 41 - Calculus Early Transcendentals

Question

Evaluate the following integrals two ways. a. Simplify the integrand first and then integrate. b. Change variables (let \(u=\ln x\) ), integrate, and then simplify your answer. Verify that both methods give the same answer. $$\int \frac{\sinh (\ln x)}{x} d x$$

Step-by-Step Solution

Answer

Question: Evaluate the integral of the given function in two ways, and verify that both methods give the same answer: $$\int \frac{\sinh (\ln x)}{x} dx$$ Answer: Both methods yield the same final result for the integral, which is $$\frac{1}{2}x + \frac{1}{2x} + C$$.

Step 1: Method 1: Simplify then Integrate

Since we are given the hyperbolic sine function, we can rewrite this using its definition with exponentials: $$\sinh(x) = \frac{e^x - e^{-x}}{2}$$ Now, let's substitute the function \(\ln x\) inside the hyperbolic sine function: $$\sinh(\ln x) = \frac{e^{\ln x} - e^{-\ln x}}{2}$$ Simplify the exponentials using properties of logarithms and exponentials: $$\sinh(\ln x) = \frac{x - \frac{1}{x}}{2}$$ Now, the integrand becomes: $$\int \frac{\sinh (\ln x)}{x}dx = \int \frac{x - \frac{1}{x}}{2x}dx = \int\frac{1}{2} - \frac{1}{2x^2}dx$$ Now, we can integrate term by term: $$\int\frac{1}{2} - \frac{1}{2x^2}dx = \frac{1}{2}\int 1 dx - \frac{1}{2}\int\frac{1}{x^2}dx$$ $$= \frac{1}{2}x - \frac{1}{2}(-x^{-1}) + C$$ The final answer for method 1 is: $$\frac{1}{2}x + \frac{1}{2x} + C$$

Step 2: Method 2: Change Variables (u-substitution)

As given in the problem statement, let \(u=\ln x\), so \(du = \frac{1}{x}dx\), and \(x = e^u\). Now substitute into the integral: $$\int \frac{\sinh (\ln x)}{x} dx = \int \frac{\sinh (u)}{e^u} e^u du$$ The \(e^u\) terms cancel out, giving: $$\int \sinh(u)du$$ Now integrate the hyperbolic sine function with respect to \(u\): $$\int \sinh(u)du = \cosh(u) + C$$ Substitute back the original variable, \(x\): $$\cosh(\ln x) + C$$ Now rewrite the hyperbolic cosine function using its definition with exponentials: $$\cosh(x) = \frac{e^x + e^{-x}}{2}$$ Substitute back the function \(\ln x\) inside the hyperbolic cosine function: $$\frac{e^{\ln x} + e^{-\ln x}}{2} + C = \frac{x + \frac{1}{x}}{2} + C$$ The final answer for method 2 is: $$\frac{1}{2}x + \frac{1}{2x} + C$$ Both methods give the same final answer, as required.