Evaluate the integrals in Exercises 41–60.
49. ∫(sech(√t)tanh(√t)dt)/√t

Verified step by step guidance

1. Start by examining the integral: \(\int \frac{\text{sech}(\sqrt{t}) \tanh(\sqrt{t})}{\sqrt{t}} \, dt\). Notice that the integrand involves \(\sqrt{t}\) inside the hyperbolic functions and also in the denominator.
2. Use the substitution method to simplify the integral. Let \(u = \sqrt{t}\), which means \(u = t^{1/2}\). Then, differentiate both sides with respect to \(t\) to find \(du\) in terms of \(dt\): \(du = \frac{1}{2\sqrt{t}} dt\) or equivalently, \(dt = 2u \, du\).
3. Rewrite the integral in terms of \(u\). Substitute \(\sqrt{t} = u\), \(dt = 2u \, du\), and replace the integrand accordingly: the denominator \(\sqrt{t}\) becomes \(u\), and the numerator remains \(\text{sech}(u) \tanh(u)\). So the integral becomes \(\int \frac{\text{sech}(u) \tanh(u)}{u} \cdot 2u \, du\).
4. Simplify the expression inside the integral. The \(u\) in the denominator and numerator cancel out, leaving \(\int 2 \text{sech}(u) \tanh(u) \, du\).
5. Recognize that the derivative of \(\text{sech}(u)\) is \(-\text{sech}(u) \tanh(u)\). Use this fact to rewrite the integral in terms of \(\text{sech}(u)\) and then integrate accordingly.

Key Concepts

Hyperbolic Functions

Hyperbolic functions like sech(x) and tanh(x) are analogs of trigonometric functions but based on hyperbolas. Understanding their definitions, properties, and derivatives is essential for manipulating and integrating expressions involving these functions.

Substitution Method in Integration

The substitution method simplifies integrals by changing variables to transform the integral into a more manageable form. Recognizing an inner function and its derivative within the integrand allows for effective substitution, especially when dealing with composite functions.

Chain Rule and Its Role in Integration

The chain rule relates the derivative of a composite function to the derivatives of its inner and outer functions. In integration, recognizing the chain rule in reverse helps identify suitable substitutions and simplifies the integral of composite functions.