Problem 56 - Single Variable Calculus

Question

find the given integral. \(\int \frac{\sinh \sqrt{x}}{\sqrt{x}} d x\)

Step-by-Step Solution

Answer

The short answer for the given integral is: \[ \int \frac{\sinh \sqrt{x}}{\sqrt{x}} d x = 2\cosh(\sqrt{x}) + C \]

Step 1: Recognize the substitution

Here, let's recognize an appropriate substitution. The most natural choice for substitution in this case is: \(u = \sqrt{x}\)

Step 2: Differentiate with respect to x

Now we need to find the differential of the chosen substitution. Differentiate "u" with respect to "x": \( \frac{du}{dx} = \frac{1}{2\sqrt{x}} \) Rearrange the equation to express "dx" in terms of "du": \( dx = 2\sqrt{x}du \)

Step 3: Substitute and simplify

Now, we will substitute "u" and "dx" into the original integral and simplify: \[ \int \frac{\sinh(u)}{u}\cdot2ud u \] The "u" in the denominator and the "2u" in the numerator cancel out. So, we are left with: \[ 2\int \sinh(u) du \]

Step 4: Integrate and substitution

Now, integrate the simplified expression with respect to "u": \[ 2\cosh(u) + C \] Finally, substitute back the original function of "u" (\(u = \sqrt{x}\)): \[ 2\cosh(\sqrt{x}) + C \]

Step 5: Write the final answer

The integral of the given function is: \[ \int \frac{\sinh \sqrt{x}}{\sqrt{x}} d x = 2\cosh(\sqrt{x}) + C \]

Key Concepts

Substitution Method

The substitution method is a useful technique in calculus for simplifying integrals. The basic idea is to make the integral simpler by substituting a part of it with a new variable. This can transform a complex integral into a more straightforward one, making it easier to solve.

To apply the substitution method, follow these steps:

• Identify a part of the integral that can be substituted with a new variable, typically to simplify the expression.
• Find the derivative of the chosen substitution with respect to the original variable.
• Rearrange this derivative to express the differential of the original variable in terms of the new variable.
• Substitute both the variable and differential expressions into the original integral.
• Simplify the new integral, if possible, and then integrate.
• Finally, substitute back the original variable to get the result in terms of the initial variable.

In the given exercise, the substitution used is \( u = \sqrt{x} \). This choice of substitution simplifies the complex expression involving \( \sinh(\sqrt{x}) \) into a more manageable form. By substituting \( u \) and rewriting \( dx \) in terms of \( du \), the integral becomes easier to evaluate.

Hyperbolic Functions

Hyperbolic functions are analogous to trigonometric functions but are based on hyperbolas instead of circles. The most common hyperbolic functions are \( \sinh \, u \) and \( \cosh \, u \). They are defined as follows:

• \( \sinh \, u = \frac{e^u - e^{-u}}{2} \)
• \( \cosh \, u = \frac{e^u + e^{-u}}{2} \)

Like their trigonometric counterparts, hyperbolic functions have excellent properties that make them useful in calculus. For example, they have derivatives:

• Derivative of \( \sinh \, u \) is \( \cosh \, u \)
• Derivative of \( \cosh \, u \) is \( \sinh \, u \)

In the exercise, we integrate \( 2 \sinh(u) \, du \) to get \( 2 \cosh(u) + C \). The relationship between \( \sinh \, u \) and its derivative, \( \cosh \, u \), directly leads us to the solution, which is elegant and straightforward. Understanding hyperbolic functions and their properties is crucial, as they commonly appear in various integrals within calculus.

Definite and Indefinite Integrals

Integrals are important tools in calculus and come in two main types: definite and indefinite. Understanding the differences and applications of each is key.

• Indefinite Integrals: These come without specific limits and represent a family of functions, generally including a constant of integration, \( C \). The result is often a general equation expressing the antiderivative.
• Definite Integrals: These have specific upper and lower limits and calculate the net area under a curve between those limits. The result of a definite integral is a number, not a function.

In the exercise provided, we're dealing with an indefinite integral: \( \int \frac{\sinh \sqrt{x}}{\sqrt{x}} \, dx \). Notice the final result includes \( C \), which accounts for any constant that might have differentiated to zero.
No limits are provided on the integral sign, indicating that we are finding a general antiderivative rather than an area under a curve between bounds. Both definite and indefinite integrals are integral parts of solving problems in calculus, each with specific roles depending on the information required from the function.