Problem 74 - Calculus Single Variable

Question

Calculate the integrals. $$ \int \frac{\exp (x)}{\sqrt{1-\exp (2 x)}} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \sin^{-1}(\exp(x)) + C \).

Step 1: Recognize the Integral Form

The integral given is \( \int \frac{\exp(x)}{\sqrt{1-\exp(2x)}} \, dx \). Notice that the denominator contains \( \sqrt{1-\exp(2x)} \) which suggests a substitution might be handy.

Step 2: Use Hyperbolic Substitution

Let's substitute \( u = \exp(x) \). Therefore, \( du = \exp(x) \, dx \) or \( dx = \frac{du}{u} \). The integral becomes \( \int \frac{1}{\sqrt{1-u^2}} \, du \).

Step 3: Recognize and Solve the New Integral Form

The transformed integral \( \int \frac{1}{\sqrt{1-u^2}} \, du \) is a standard inverse trigonometric integral which is known to be \( \sin^{-1}(u) + C \).

Step 4: Substitute Back

Recall \( u = \exp(x) \). Substitute back to get the original variables in the solution: \( \sin^{-1}(\exp(x)) + C \).

Step 5: Final Solution

Thus, the integral \( \int \frac{\exp(x)}{\sqrt{1-\exp(2x)}} \, dx \) evaluates to \( \sin^{-1}(\exp(x)) + C \).

Key Concepts

Integration Techniques

When facing an integral that seems complex at first glance, like the one presented here, using various **integration techniques** can simplify the process. The integral given is:\[ \int \frac{\exp(x)}{\sqrt{1-\exp(2x)}} \ dx \]To tackle this, notice the combination of exponential and square root terms. Here are some general strategies for similar problems:

• Identify substitutions that simplify the integrand.
• Use known identities and transformations to match the integrand to a familiar form, often found in standard integral tables.
• Examine parts of the integral that resemble trigonometric or hyperbolic forms.

By using hyperbolic substitution, as outlined in the next section, we reduce the complexity and recognize a form that leads us to an inverse trigonometric function.

Hyperbolic Substitution

In mathematics, **hyperbolic substitution** is an extremely useful technique for transforming complex integrals into simpler forms. In the exercise, the substituted variable \( u = \exp(x) \) simplifies the expression to a form that is easier to handle.This method works exceptionally well with integrands that contain forms like \( \sqrt{1-u^2} \) or \( \sqrt{u^2-1} \). Here are steps to consider while performing hyperbolic substitution:

• Choose a substitution that transforms the complicated part of the integrand (typically under the square root) into a standard quadratic form.
• Substitute and rewrite the integral in terms of the new variable.
• Evaluate the integral by recognizing the form as a standard integral, such as that of inverse trigonometric functions.
• Substitute back to express the solution in terms of the original variable.

Hyperbolic substitution often transforms the problem into a simpler calculation, revealing trigonometric relationships that might not be immediately obvious.

Inverse Trigonometric Functions

The core of solving this integral relies on recognizing the appearance of **inverse trigonometric functions**. Once we make the substitution \( u = \exp(x) \), the integral simplifies to\[ \int \frac{1}{\sqrt{1-u^2}} \, du \]This is a standard form that corresponds to the inverse sine function, \( \sin^{-1}(u) \). Here are some important aspects:

• Inverse trigonometric integrals arise when you integrate functions that involve expressions like \( \frac{1}{\sqrt{a^2 - u^2}} \).
• These functions include \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), and \( \tan^{-1}(x) \), amongst others.
• Recognizing the specific form is key to applying the correct inverse function directly, often found in integral tables.

After solving for the integral using the inverse sine, it's crucial to substitute back the original expression \( \exp(x) \) to complete the process. The final result is elegantly simple due to these inverse relationships.