Problem 38 - Calculus Early Transcendentals

Question

Evaluate each definite integral. $$\int_{0}^{4} \frac{\operatorname{sech}^{2} \sqrt{x}}{\sqrt{x}} d x$$

Step-by-Step Solution

Answer

Question: Evaluate the definite integral $$\int_{0}^{4} \frac{\operatorname{sech}^2\sqrt{x}}{\sqrt{x}} dx.$$ Answer: The definite integral evaluates to $$- 4\tanh(2) + 2\ln(\cosh(2)).$$

Step 1: Substitution

Let \(u = \sqrt{x}\). Then, $$ x = u^2 \Rightarrow d x = 2u \, d u $$ Now, let's change the limits of integration. Since the integral is definite, we should update the limits as well. When \(x = 0\), \(u = \sqrt{0} = 0\). When \(x=4\), \(u=\sqrt{4}=2\). So, the limits of integration are changed from \(x=0\) to \(x=4\) to \(u=0\) to \(u=2\). After the substitution and changing the limits, we get a new integral, $$ \int_{0}^{2} \operatorname{sech}^2 u (2u) du. $$

Step 2: Evaluate the integral

Now, we need to evaluate the resulting integral, $$ \int_{0}^{2} 2u\operatorname{sech}^2u du. $$ To solve this integral, we will make use integration by parts formula: $$ \int_a^b u(v)dv = u(b)v(b)-u(a)v(a)-\int_a^b v(du). $$ We will choose \(w=u\) and its derivative \(d w = d u\), and \(d v = 2 \operatorname{sech}^2u du\) for which we will find the antiderivative \(v\). We know that antiderivative of \(2\operatorname{sech}^2(u)\) is \(-2\tanh(u)\). Therefore, $$ v = -2\tanh(u). $$ Now let's apply integration by parts: \begin{align*} \int_{0}^{2} u \cdot 2\operatorname{sech}^2u du &= -\left[u\cdot 2\tanh(u)\right]_0^2 + 2\int_{0}^{2} \tanh(u) du \\ &= - \left[2\cdot 2\tanh(2)- 0\right] + 2\left[\ln(\cosh(u))\right]_0^2 . \end{align*}

Step 3: Calculate the result

Now, let's find the result of the definite integral: \begin{align*} & - \left[2\cdot 2\tanh(2)- 0\right] + 2\left[\ln(\cosh(u))\right]_0^2 \\ &= - 4\tanh(2) + 2\ln(\cosh(2)) - 2\ln(\cosh(0)) \\ &= - 4\tanh(2) + 2\ln(\cosh(2)) - 2\ln(1) \\ & = - 4\tanh(2) + 2\ln(\cosh(2)). \end{align*} Therefore, the solution to the given integral is $$ \int_{0}^{4} \frac{\operatorname{sech}^2\sqrt{x}}{\sqrt{x}} dx = - 4\tanh(2) + 2\ln(\cosh(2)). $$

Key Concepts

Integration by Substitution

Integration by substitution is a handy technique used in calculus to simplify the process of integrating complex functions. It's akin to reversing the chain rule from differentiation.
For instance, consider the function within our initial integral: \( \frac{\operatorname{sech}^2 \sqrt{x}}{\sqrt{x}} \). By replacing \( \sqrt{x} \) with \( u \) (i.e., \( u = \sqrt{x} \)), the derivative \( du = \frac{1}{2\sqrt{x}} dx \) can be rearranged to express \( dx = 2u \, du \), simplifying the original integral.
This substitution essentially changes the expression into new limits that work over \( u \) instead of \( x \), which leads us to a more manageable integral.

• This process involves several steps starting from the substitution for integration itself,
• changing the integration limits due to switching variables,
• and finally simplifying the integrand.

By following these steps, we convert an otherwise daunting integral into one that can be tackled with more straightforward techniques, such as integration by parts.

Integration by Parts

Integration by parts is another crucial technique in calculus, derived from the product rule for differentiation. It allows us to solve integrals involving products of functions. The formula is given by:# \( \int u \, dv = uv - \int v \, du \).
In our problem, after substituting, we arrive at the integral \( \int 2u \, \operatorname{sech}^2 u \, du \). To apply integration by parts here, we set \( u = u \) and \( dv = 2\operatorname{sech}^2 u \, du \). Consequently, by integrating \( dv \), we get \( v = -2\tanh(u) \).
This allows us to reframe the problem using the integration by parts formula:

• Multiply \( u \) with \( v \), which becomes \(-2u\tanh(u)\),
• Evaluate this expression at the updated limits \( 0 \) and \( 2 \),
• And deal with the remaining integral \( \int \tanh(u) \, du \).

This method effectively breaks down the integral into simpler parts that can be solved systematically, and it's especially useful when one part of the integrand is a straightforward antiderivative.

Hyperbolic Functions

Hyperbolic functions appear frequently in integration problems, offering analogs to trigonometric functions but with distinct properties. Key hyperbolic functions include \( \sinh \), \( \cosh \), and \( \tanh \). These functions can be defined in terms of exponential functions:

• \( \sinh(x)=\frac{e^x-e^{-x}}{2} \),
• \( \cosh(x)=\frac{e^x+e^{-x}}{2} \), and
• \( \tanh(x)=\frac{\sinh(x)}{\cosh(x)} \).

In the context of our problem, the function \( \operatorname{sech}^2 u \) or \( \frac{1}{\cosh^2 u} \) is the derivative of \( \tanh(u) \). Thus, knowing the derivatives and integrals of hyperbolic functions is extremely useful.
Hyperbolic functions excel in expressing certain geometric properties and equations, particularly in physics and engineering fields. Within integration, they provide elegant solutions to what might otherwise be complicated rational function problems. Their behavior is predictable, similar to trigonometric identities, simplifying finding antiderivatives and derivatives alike.