Question

Indefinite integrals Use a change of variables or Table 5.6 to evaluate the following indefinite integrals. Check your work by differentiating. $$\int \frac{e^{w}}{36+e^{2 w}} d w$$

Short answer

In the given integral, $$\int \frac{e^{w}}{36+e^{2 w}} d w$$, we used substitution, choosing $$u = e^w $$, to simplify the expression and solve the integral. The final solution was found to be: $$\int \frac{e^{w}}{36+e^{2 w}} d w = \frac{1}{12} \ln \left(\frac{e^{2w}}{36 + e^{2w}}\right) + C$$ The result was confirmed to be correct by differentiating the solution and obtaining the original integrand, $$\frac{e^w}{36+e^{2 w}}$$.

Step-by-step solution

Choose a substitution

Let us choose the substitution $$u=e^w$$, then $$du = e^w \, dw$$. Now we will rewrite the given integral in terms of $$u$$.

Rewrite the integral

Substitute $$u$$ and $$du$$ into the integral: $$\int \frac{e^{w}}{36+e^{2 w}} d w = \int \frac{u}{36+u^2} du$$ We now have an integral that is easier to evaluate.

Solve the integral

This is a standard integral in the form of $$\int \frac{u}{a^2 + u^2} du$$, which can be solved directly using the formula: $$\int \frac{u}{a^2 + u^2} du = \frac{1}{2a} \ln \left(\frac{u^2}{a^2 + u^2}\right) + C$$, where $$a$$ is a constant and $$C$$ is the constant of integration. In our case, $$a^2 = 36$$, so $$a = 6$$. Applying this formula, we get: $$\int \frac{u}{36+u^2} du = \frac{1}{2\cdot6} \ln \left(\frac{u^2}{36 + u^2}\right) + C$$ Simplify the expression to obtain: $$\int \frac{u}{36+u^2} du = \frac{1}{12} \ln \left(\frac{u^2}{36 + u^2}\right) + C$$

Resubstitute

Now, resubstitute $$u=e^w$$ back into the expression to find the original solution for the given integral: $$\int \frac{e^{w}}{36+e^{2 w}} d w = \frac{1}{12} \ln \left(\frac{e^{2w}}{36 + e^{2w}}\right) + C$$ So, the solution to the given integral is: $$\int \frac{e^{w}}{36+e^{2 w}} d w = \frac{1}{12} \ln \left(\frac{e^{2w}}{36 + e^{2w}}\right) + C$$

Check the work by differentiating

To check our work, differentiate the obtained solution with respect to $$w$$: $$\frac{d}{dw}\left(\frac{1}{12} \ln \left(\frac{e^{2w}}{36 + e^{2w}}\right) + C\right)$$ Applying the chain rule, we have: $$\frac{e^{2w}(2)(36+e^{2w}) - e^{2w}(2e^{2w})}{12(36 + e^{2w})^2}$$ Simplify the expression: $$\frac{2e^{2w}(36+e^{2w}-e^{2w})}{12(36 + e^{2w})^2} = \frac{e^w(72)}{12(36 + e^{2w})^2}$$ So, we end up with: $$\frac{e^w}{36+e^{2 w}}$$ Since our differentiation result matches the original integrand, this confirms that our solution is correct.

Key concepts

Integration by Substitution

Integration by substitution is a fundamental technique used to simplify the process of finding indefinite integrals. The method involves replacing the original variable in the integrand with a new variable, making integration more straightforward.

Here's a step-by-step approach to using substitution:

• Identify the inner function within the integrand that can be substituted to simplify the integral.
• Choose a substitution where the derivative of the new variable is also present in the integrand.
• Rewrite the entire integral in terms of the new variable and its differential.
• Integrate with respect to the new variable.
• Finally, replace the substituted variable back with the original variable to complete the solution.

Substitution not only makes the integral easier but can often convert it into a basic form that can be solved using standard formulas or direct integration.

Natural Logarithm Properties

The natural logarithm, denoted as \( \ln(x) \), has properties that frequently prove useful when integrating functions involving logarithms. Some of these properties are:

• \( \ln(ab) = \ln(a) + \ln(b) \), which is the property of logarithms that states that the log of a product is equal to the sum of the logs.
• \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \), indicating the natural log of a quotient is the difference of the natural logs.
• \( \ln(a^p) = p\ln(a) \), which is the power rule, showing that the logarithm of a power is the power of the logarithm.

These properties are particularly helpful when the integrand is a fractional expression and can pave the way for a more straightforward integration process after substitution.

U-Substitution Method

The u-substitution method is an application of the integration by substitution technique. When using u-substitution:

• Select a part of the integrand to be \( u \), typically the inner function of a composite function.
• Determine \( du \), the derivative of \( u \).
• Ensure that \( du \), or a scalar multiple of it, is present in the integral to allow for a total substitution.
• Convert the integral entirely into terms of \( u \) and \( du \), and then carry out the integration.
• Substitute back the original variables to express the antiderivative in the original terms.

When done correctly, this method transforms a complex integral into one that is easier to solve, often one that matches a known integral form. It is particularly valuable when integrating rational functions or functions involving square roots, trigonometric functions, or exponentials.

Integration Techniques

There are several powerful techniques for solving integrals, and a deep understanding of these methods is essential for tackling a wide range of problems:

• Standard Integration: Integrating standard functions using known formulas.
• Integration by Substitution: Simplifying integrals through variable substitution.
• Integration by Parts: Based on the product rule for differentiation, it breaks down integrals involving products of functions.
• Partial Fractions Decomposition: Used for integrating rational functions by breaking them into simpler fractions.
• Trigonometric Substitution: Useful for integrals involving square roots of quadratic expressions or trigonometric identities.

These techniques can be used independently or in combination to solve complex integrals. It is often the case that the choice of the appropriate method significantly impacts the ease of finding an integral’s solution. An understanding of the particular intricacies of each technique, along with practice, allows for a more efficient and successful application.