Problem 55
Question
find the given integral. \(\int \cosh (2 x+3) d x\)
Step-by-Step Solution
Verified Answer
The short answer for the given integral is: \(\int \cosh(2x + 3) dx = \frac{1}{2} \sinh(2x + 3) + C\).
1Step 1: Substitution
Let's make the substitution:
\(u = 2x + 3\)
Now, we find the derivative of u with respect to x:
\(\frac{du}{dx} = 2\)
Rearrange to find dx:
\(dx = \frac{1}{2} du\)
Now, replace dx and (2x+3) with their respective expressions in terms of u:
\(\int \cosh(u) \frac{1}{2} du\)
2Step 2: Integrate
Next, we integrate with respect to u:
\(\frac{1}{2} \int \cosh(u) du\)
The integral of the hyperbolic cosine function is simply the hyperbolic sine function:
\(\frac{1}{2} \sinh(u) + C\), where C is the constant of integration.
3Step 3: Substitute back
Finally, we will substitute the expression for u back in terms of x:
\(\frac{1}{2} \sinh(2x + 3) + C\)
So, the integral of the given function is:
\(\int \cosh(2x + 3) dx = \frac{1}{2} \sinh(2x + 3) + C\)
Key Concepts
Substitution MethodHyperbolic FunctionsIntegration Techniques
Substitution Method
The substitution method is a popular technique used in calculus to simplify integrals, especially when dealing with functions that might not be straightforward to integrate in their current form. This technique involves substituting part of the integrand with a new variable, typically denoted as \(u\), to make the integration process simpler.
In the given problem, we have the integral \(\int \cosh(2x + 3) \, dx\). The expression \(2x + 3\) inside the hyperbolic cosine is the complex part of the integrand. We set \(u = 2x + 3\), effectively treating it as a new variable. This transformation is designed to reduce the complexity of the integral. Performing this substitution means we also need the derivative of \(u\) with respect to \(x\): \(\frac{du}{dx} = 2\). Solving for \(dx\), we get \(dx = \frac{1}{2} du\).
After substitution, the integral changes to \(\int \cosh(u) \frac{1}{2} du\). This makes the problem much easier to handle, as we’re now dealing with a basic hyperbolic cosine function in terms of \(u\). Once integrated, we substitute back to the original variable \(x\) to find the solution to the original integral.
In the given problem, we have the integral \(\int \cosh(2x + 3) \, dx\). The expression \(2x + 3\) inside the hyperbolic cosine is the complex part of the integrand. We set \(u = 2x + 3\), effectively treating it as a new variable. This transformation is designed to reduce the complexity of the integral. Performing this substitution means we also need the derivative of \(u\) with respect to \(x\): \(\frac{du}{dx} = 2\). Solving for \(dx\), we get \(dx = \frac{1}{2} du\).
After substitution, the integral changes to \(\int \cosh(u) \frac{1}{2} du\). This makes the problem much easier to handle, as we’re now dealing with a basic hyperbolic cosine function in terms of \(u\). Once integrated, we substitute back to the original variable \(x\) to find the solution to the original integral.
Hyperbolic Functions
Hyperbolic functions, much like their trigonometric siblings, are essential in many calculus problems. These functions, typically including hyperbolic sine (\(\sinh\)) and hyperbolic cosine (\(\cosh\)), appear frequently in integrals due to their unique characteristics related to exponential functions.
The hyperbolic cosine function, \(\cosh(x)\), is defined by the expression \(\cosh(x) = \frac{e^x + e^{-x}}{2}\). It shares several properties with the cosine function, for instance, \(\cosh(0) = 1\), although it doesn't oscillate like the cosine does. Instead, \(\cosh(x)\) is an even function that never goes negative.
In integration, knowing the derivative of \(\cosh(x)\), which is \(\sinh(x)\), is crucial. This property is used in the substitution technique and directly affects how we simplify the integral. Understanding hyperbolic functions and their integrals allows students to tackle a variety of calculus problems in physics and engineering.
The hyperbolic cosine function, \(\cosh(x)\), is defined by the expression \(\cosh(x) = \frac{e^x + e^{-x}}{2}\). It shares several properties with the cosine function, for instance, \(\cosh(0) = 1\), although it doesn't oscillate like the cosine does. Instead, \(\cosh(x)\) is an even function that never goes negative.
In integration, knowing the derivative of \(\cosh(x)\), which is \(\sinh(x)\), is crucial. This property is used in the substitution technique and directly affects how we simplify the integral. Understanding hyperbolic functions and their integrals allows students to tackle a variety of calculus problems in physics and engineering.
Integration Techniques
Integration techniques are strategies used to solve integrals, which can range from simple to extremely complex. Among these techniques, substitution, integration by parts, and partial fraction decomposition are widely used based on the structure of the integral.
In solving \(\int \cosh(2x + 3) \, dx\), the substitution method is applied effectively. Choosing the substitution \(u = 2x + 3\) reduces it to a simpler integral involving \(\cosh(u)\). This choice aligns with the structure since the derivative \(\frac{d}{dx}(2x + 3)\) is constant, a condition that makes substitution particularly effective.
Once substituted, the integral \(\int \cosh(u) \, du\) is straightforward because the antiderivative of \(\cosh(u)\) is \(\sinh(u)\). This innate property of hyperbolic functions makes them much simpler to integrate under such transformations.
Different integration methods leverage different techniques. Substitution leverages changes of variable to simplify integrands, whereas other techniques may rearrange or decompose integrands' forms to achieve integrability. Recognizing when to use each method can greatly enhance one's problem-solving efficiency in calculus.
In solving \(\int \cosh(2x + 3) \, dx\), the substitution method is applied effectively. Choosing the substitution \(u = 2x + 3\) reduces it to a simpler integral involving \(\cosh(u)\). This choice aligns with the structure since the derivative \(\frac{d}{dx}(2x + 3)\) is constant, a condition that makes substitution particularly effective.
Once substituted, the integral \(\int \cosh(u) \, du\) is straightforward because the antiderivative of \(\cosh(u)\) is \(\sinh(u)\). This innate property of hyperbolic functions makes them much simpler to integrate under such transformations.
Different integration methods leverage different techniques. Substitution leverages changes of variable to simplify integrands, whereas other techniques may rearrange or decompose integrands' forms to achieve integrability. Recognizing when to use each method can greatly enhance one's problem-solving efficiency in calculus.
Other exercises in this chapter
Problem 54
find the derivative of the function. \(y=2 x \operatorname{coth}^{-1} 2 x-\ln \sqrt{1-4 x^{2}}\)
View solution Problem 54
Verify the formula for the volume of a sphere of radius \(r\) by finding the volume of the solid obtained by revolving the region bounded by the graph of \(x^{2
View solution Problem 55
Flight Path of an Airplane The path of an airplane on its final approach to landing is described by the equation \(y=f(x)\) with $$ \begin{array}{r} f(x)=4.3403
View solution Problem 55
Find the volume of a cap of height \(h\) formed from a sphere of radius \(r\).
View solution