Problem 43 - Calculus: One and Several Variables

Question

Integrate by setting \(u=\tanh \frac{1}{2} x.\) $$\int \frac{1}{\sinh x+\cosh x} d x$$

Step-by-Step Solution

Answer

The short answer is: \[\int \frac{1}{\sinh x+\cosh x} d x = \frac{1}{2}\sqrt{1+\tanh^2\cdot \frac{1}{2}x}+C\]

Step 1: Define u and find its derivative with respect to x

Let \(u = \tanh(\frac{1}{2}x)\). To find the derivative of \(u\) with respect to \(x\), we have: \[\frac{du}{dx} = \frac{1}{2} \cdot \text{sech}^2\left(\frac{1}{2}x\right)\]

Step 2: Find dx in terms of du

Now, from the derivative \(\frac{du}{dx}\), we will find \(dx\) in terms of \(du\): \[dx = 2 \cdot \text{sech}^2 \left(\frac{1}{2}x\right) du \]

Step 3: Rewrite the integral in terms of u using the relation between u and x

Rewriting the integral in terms of \(u\), start by observing that: \[\sinh x + \cosh x = \dfrac{e^x + e^{-x} + e^x - e^{-x}}{2} = \dfrac{2 e^x}{2} = e^x\] Now, using the hyperbolic tangent definition and the relation between hyperbolic functions: \[\tanh(z) = \dfrac{\sinh z}{\cosh z} \Rightarrow \sinh z = \tanh (z)\cdot \cosh(z)\] Applying this result to our \(u\): \[\sinh x = \tanh\left(\frac{1}{2}x\right) \cosh\left(\frac{1}{2}x\right) = u \cosh\left(\frac{1}{2}x\right)\] Using the relation \(\cosh^2(z) - \sinh^2(z) = 1\) and the previous result: \[\cosh^2\left(\frac{1}{2}x\right) - u^2 = 1 \Rightarrow \cosh\left(\frac{1}{2}x\right) = \sqrt{1+u^2}\] Taking the derivative of last expression with respect to x: \[\frac{1}{2}\cosh\left(\frac{1}{2}x\right)\cdot\frac{du}{dx} = \frac{d}{dx}\sqrt{1+u^2}\] And using the definition of derivative for u: \[\text{sech}^2\left(\frac{1}{2}x\right) = \frac{2}{\cosh^2\left(\frac{1}{2}x\right)}\] Substituting dx, the integral will be rewritten as: \[\int \frac{1}{e^x} \cdot 2 \cdot \text{sech}^2\left(\frac{1}{2}x\right) du = \int \frac{1}{u\sqrt{1+u^2}}du\]

Step 4: Solve the integral in terms of u

Now, we can solve the integral in terms of \(u\): \[\int \frac{1}{u\sqrt{1+u^2}}du = \int \frac{1}{u\sqrt{(1+u^2)}}du\] Since \(\frac{d}{du}(1+u^2) = 2u\), we can perform a simple substitution: \[ \int \frac{1}{u\sqrt{1+u^2}}du = \frac{1}{2}\int \frac{1}{\sqrt{1+u^2}} (2u) du = \frac{1}{2} \int \frac{d(1+u^2)}{\sqrt{1+u^2}}\] Now, integrating: \[\frac{1}{2} \int \frac{d(1+u^2)}{\sqrt{1+u^2}} = \frac{1}{2}\sqrt{1 +u^2} + C\]

Step 5: Substitute back x to get the final solution

Finally, we substitute back the original variable \(x\) using the relation \(u = \tanh(\frac{1}{2}x)\): \[\frac{1}{2}\sqrt{1 + \tanh^2\left(\frac{1}{2}x\right)} + C = \frac{1}{2}\sqrt{1+\tanh^2\cdot \frac{1}{2}x}+C\]

Key Concepts

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but for a hyperbola instead of a circle. They include hyperbolic sine (\(\sinh(x)\)) and hyperbolic cosine (\(\cosh(x)\)), which can be defined in terms of exponential functions:

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

The identity \(\cosh^2(x) - \sinh^2(x) = 1\) resembles the Pythagorean identity for sine and cosine.
\(\tanh(x)\) is another hyperbolic function given by the ratio \(\frac{\sinh(x)}{\cosh(x)}\), similar to \(\tan(\theta)\) in trigonometry.​ These functions are crucial in various areas like calculus, physics, and engineering.
If you understand these identities, along with their representations, they become valuable tools for solving integrals involving hyperbolic functions.

Derivative of Hyperbolic Tangent

The derivative of a function gives us insights into its rate of change. For the hyperbolic tangent function \(\tanh(x)\), this derivative is particularly interesting.

• The derivative of \(\tanh(x)\) with respect to \(x\) is given by \(\text{sech}^2(x)\).

This can be understood better if we recall the identity for hyperbolic functions: \(\text{sech}^2(x) = 1 - \tanh^2(x)\).
When substituting variables in integrals, knowing the derivatives of functions allows for proper transformations.
For example, to use \(u = \tanh\left(\frac{1}{2}x\right)\), the derivative helps find the differential \(dx\) in terms of \(du\). This is critical for solving integrals such as the one presented in the problem.

Trigonometric Substitution

Trigonometric substitution involves replacing variables in integrals with trigonometric identities to simplify them. This technique often helps with integrals involving squares or square roots.
In the context of integrating hyperbolic functions, this method can also be convenient. We can use identities from trigonometry and hyperbolic functions to simplify the integration process.

• For an integral with \(\sqrt{1+u^2}\), a substitution like \(u = \tanh\left(z\right)\) is handy.
• Trigonometric substitution helps by transforming complex expressions into easier-to-handle forms.

While trigonometric substitution is common with circular functions, its principles extend smoothly to hyperbolic functions, broadening the realm of calculus applications and problem-solving strategies.

Integration Techniques

The field of integration encompasses various techniques to solve complex integrals. Among them, substitution is a powerful tool.

• Substitution helps simplify integrals by changing variables to their simpler forms.
• When handling sums or products of hyperbolic functions, substitution can turn the integral into a more recognizable one for easy calculation.

In solving the provided integral, substitution allowed us to transform the integral involving hyperbolic functions into one solvable with familiar methods.
Other techniques may include integration by parts or partial fractions, but being comfortable with substitution can unlock easier solutions to seemingly challenging integrals, particularly those involving hyperbolic or trigonometric functions.