Question

$$\text {Evaluate the following integrals.}$$ $$\int e^{x} \sec \left(e^{x}+1\right) d x$$

Short answer

Question: Evaluate the integral of the function $$e^x \sec(e^x + 1)$$. Answer: The integral cannot be expressed in terms of elementary functions. However, we can partially evaluate it as follows: $$\int e^x \sec\left(e^x+1\right) dx = \ln(e^x) \ln|\sec(e^x + 1) + \tan(e^x + 1)| - \int \ln|\sec(e^x+1)+\tan(e^x+1)| dx + C$$.

Step-by-step solution

1. Identify the substitution

Given the integrand $$e^x \sec(e^x + 1)$$, we will use substitution to make this integral easier to evaluate. The most natural substitution seems to be setting $$u = e^x + 1$$, which would transform the integral into something more manageable.

2. Determine the differential

In order to perform the substitution, we will need to find the differential $$du$$. To do that, we differentiate $$u = e^x + 1$$ with respect to $$x$$: $$\frac{du}{dx} = \frac{d(e^x + 1)}{dx} = e^x$$. Now, we can solve for $$dx$$: $$dx = \frac{du}{e^x}$$.

3. Substitute and simplify the integral

Now we can substitute our expressions for $$u$$ and $$dx$$ into the integral: $$\int e^x \sec(e^x + 1) dx = \int \sec(u) \frac{du}{e^x}$$. Since $$u=e^x+1$$, we can solve for $$e^x$$: $$e^x = u-1$$. Thus, the integral simplifies to: $$\int \sec(u) \frac{du}{u-1}$$.

4. Integration by parts

To evaluate the remaining integral, we will use integration by parts. We'll let $$dv = \sec(u) du$$ and $$u = \ln(u - 1)$$: $$\frac{d}{du}v=\sec(u) \Rightarrow v = \int sec(u) du = \ln|\sec(u) + \tan(u)| + C$$, and $$\frac{d}{du}u = \frac{1}{u-1} \Rightarrow du=(u-1) \cdot du$$. Using integration by parts formula: $$\int u dv = uv - \int v du$$, we get: $$\int \ln(u-1) \sec(u) du = \ln(u-1) \ln|\sec(u) + \tan(u)| - \int \ln|\sec(u)+\tan(u)| du$$.

5. Back substitution

Now we need to substitute back $$x$$ in place of $$u$$: $$\int e^x \sec\left(e^x+1\right) dx = \ln(e^x) \ln|\sec(e^x + 1) + \tan(e^x + 1)| - \int \ln|\sec(e^x+1)+\tan(e^x+1)| dx + C$$. Unfortunately, we can't simplify the remaining integral further. Nevertheless, we have made some progress in expressing the integral in terms of simpler functions. The final answer is: $$\int e^x \sec\left(e^x+1\right) dx = \ln(e^x) \ln|\sec(e^x + 1) + \tan(e^x + 1)| - \int \ln|\sec(e^x+1)+\tan(e^x+1)| dx + C$$.

Key concepts

Substitution Method

Integration can often be simplified using the Substitution Method. This technique replaces a complex piece of the integrand with a simpler variable, making the integral more manageable. To choose a substitution:

• Identify a suitable substitution to replace part of the integrand. Usually, a function inside the integrand that has a simple derivative is a good candidate. In our exercise, we selected \( u = e^x + 1 \).
• Calculate the derivative \( du \) to know what to replace \( dx \) with in the integral. Here, we found that \( du = e^x \, dx \).
• Express the entire integral with the new variable, including re-expressing any remaining parts in terms of \( u \).

This method reduces the range of the problem, making it simpler to solve compared to the original form.

Integration by Parts

When substitution alone doesn't simplify the integral sufficiently, Integration by Parts can be a powerful technique. It is particularly handy for products of functions that do not simplify easily with substitution.

• This method uses the formula \( \int u \, dv = uv - \int v \, du \), translating the integral into parts that might be easier to handle individually. This formula is inspired by the product rule in differentiation.
• Choose \( u \) and \( dv \) from the integrand such that \( du \) and \( v \) are straightforward to compute. In our problem, we selected \( dv = \sec(u) \, du \) and \( u = \ln(u-1) \).
• After applying the formula, any remaining integrals may still require further simplification or different techniques.

Integration by Parts can be repeated multiple times if the resulting integrals necessitate it. This flexibility makes it a valuable technique for evaluating complex integrals.

Differential Calculus

Differential Calculus plays a critical role in integration techniques like substitution and integration by parts. Understanding how to differentiate functions allows us to reverse the process, finding antiderivatives or integrals.

• Key operations involve finding the derivative of functions. In substitution, you differentiate to find \( du \), as we did by differentiating \( u = e^x + 1 \) to get \( du = e^x \, dx \).
• In integration by parts, calculating \( du \) for the chosen \( u \) is essential. Understanding the differentiation rules is foundational here.
• Knowledge of derivatives of basic and complex functions speeds up the integration process and accuracy.

Mastering differential calculus is essential, as it bridges many integration techniques by identifying strategies to make integration more feasible.

Trigonometric Functions

Trigonometric Functions like sine, cosine, and secant often appear in integrals, demanding specific knowledge for successful evaluation. Understanding these functions and their properties can clarify the integration process.

• Commonly seen in integrands, trigonometric function integrals challenge students due to their cyclical nature and intricate derivatives.
• Key identities and derivatives of trigonometric functions should be memorized for efficient problem-solving. For example, the secant function's derivative is frequently encountered in problems.
  The integral of \( \sec(u) \) was a crucial component in this exercise.
• Using fundamental identities, converting between trigonometric forms can sometimes simplify the integration further.

Recognizing when and how to apply these identities is a powerful skill, particularly when the integral includes complex expressions involving trigonometric functions.