Problem 7 - Active Calculus

Question

For each of the following integrals involving radical functions, (1) use an appropriate \(u\) -substitution along with Appendix A to evaluate the integral without the assistance of technology, and (2) use a CAS to evaluate the original integral to test and compare your result in (1). a. \(\int \frac{1}{x \sqrt{9 x^{2}+25}} d x\) b. \(\int x \sqrt{1+x^{4}} d x\) c. \(\int e^{x} \sqrt{4+e^{2 x}} d x\) d. \(\int \frac{\tan (x)}{\sqrt{9-\cos ^{2}(x)}} d x\)

Step-by-Step Solution

Answer

a. \( \frac{1}{9}\sqrt{9x^2+25} + C \), b. \( \frac{1}{6}(1+x^4)^{3/2} + C \), c. \( \frac{1}{3}(4+e^{2x})^{3/2} + C \), d. \( - \sin^{-1}(\frac{\cos(x)}{3}) + C \).

Step 1: Problem (a) - Identify the substitution

Consider the integral \(\textstyle \textstyle \int \frac{1}{x \sqrt{9x^2+25}}dx \). Let's use substitution \(\textstyle u=9x^2+25 \). Then \(\textstyle du = 18x \, dx \) or \(\textstyle dx=\frac{du}{18x} \).

Step 2: Rewrite in terms of u

Using the substitution \(\textstyle u=9x^2+25 \) in the integral, we can rewrite it as \(\textstyle \int \frac{1}{x \sqrt{u}} \cdot \frac{du}{18x} \). Simplifying this, we get \(\textstyle \frac{1}{18} \int \frac{1}{u^{\frac{1}{2}}} du \).

Step 3: Integrate

Integrate \(\textstyle \frac{1}{18} \int u^{-\frac{1}{2}} du \). Using the power rule we get \(\textstyle \frac{1}{18} \cdot 2u^{\frac{1}{2}} = \frac{2}{18}u^{\frac{1}{2}} \). Simplifying this gives \(\textstyle \frac{1}{9} \sqrt{u} \).

Step 4: Substitute back to x

Finally, substituting back for \(\textstyle u = 9x^2 + 25 \), we have \(\textstyle \frac{1}{9}\sqrt{9x^2+25} \).

Step 5: Confirmation using CAS

Use a Computer Algebra System (CAS) to evaluate the original integral and confirm that \(\textstyle \int \frac{1}{x \sqrt{9x^2+25}}dx = \frac{1}{9}\sqrt{9x^2+25} + C \).

Step 6: Problem (b) - Identify the substitution

Consider the integral \(\textstyle \int x \sqrt{1+x^4} \, dx \). Let's use substitution \(\textstyle u = 1 + x^4 \). Then \(\textstyle du = 4x^3 \, dx \). However, to facilitate the substitution, split \(\textstyle x \, dx \) separately.

Step 7: Rewrite in terms of u

Rewrite the integral \(\textstyle \int x \sqrt{1+x^4}dx \) as \(\textstyle \frac{1}{4} \int \sqrt{u} du \) because \(\textstyle dx = \frac{du}{4x^3} \). Here, remembering that if we split, the bounds also simplify.

Step 8: Integrate

Integrate \(\textstyle \frac{1}{4} \int u^{\frac{1}{2}} du \). Using the power rule for integration, we get \(\textstyle \frac{1}{4} \cdot \frac{2}{3} u^{3/2} = \frac{1}{6}u^{3/2} \).

Step 9: Substitute back to x

Finally, substituting back for \(\textstyle u = 1 + x^4 \), we obtain \(\textstyle \frac{1}{6}(1+x^4)^{3/2} + C \).

Step 10: Confirmation using CAS

Use a Computer Algebra System (CAS) to evaluate the original integral and confirm that \(\textstyle \int x \sqrt{1+x^4}dx = \frac{1}{6}(1+x^4)^{3/2} + C \).

Step 11: Problem (c) - Identify the substitution

Consider the integral \(\textstyle \int e^{x} \sqrt{4+e^{2x}}dx \). Let's use substitution \(\textstyle u = 4+e^{2x} \). Then \(\textstyle du = 2e^{2x} dx \) or \(\textstyle dx = \frac{du}{2e^{2x}} \).

Step 12: Rewrite in terms of u

Using the substitution \(\textstyle u = 4 + e^{2x} \) in the integral, rewrite it as \(\textstyle \int \frac{e^{x}}{\sqrt{u}} \cdot \frac{du}{2e^{2x}} \). Simplifying this, we get \(\textstyle \frac{1}{2} \int \frac{1}{\sqrt{u}}e^{-x}du \). Noticing that \(\textstyle e^{-x} = \sqrt{\frac{u}{4}} \).

Step 13: Integrate

Integrate \(\textstyle \frac{1}{2} \int u^{\frac{1}{2}} du \). Using the power rule for integration, we get \(\textstyle \frac{1}{2} \cdot \frac{2}{3} u^{3/2} = \frac{1}{3}u^{3/2} \).

Step 14: Substitute back to x

Finally, substituting back for \(\textstyle u = 4+e^{2x} \), we obtain \(\textstyle \frac{1}{3}(4+e^{2x})^{3/2} + C \).

Step 15: Confirmation using CAS

Use a Computer Algebra System (CAS) to evaluate the original integral and confirm that \(\textstyle \int e^{x} \sqrt{4 + e^{2x}}dx = \frac{1}{3}(4+e^{2x})^{3/2} + C \).

Step 16: Problem (d) - Identify the substitution

Consider the integral \(\textstyle \int \frac{\tan(x)}{\sqrt{9-\cos^2(x)}}dx \). Let's use substitution \(\textstyle u = \cos(x) \). Then \(\textstyle du = -\sin(x) \, dx \) or \(\textstyle - du = \sin(x) \, dx \).

Step 17: Rewrite in terms of u

Using the substitution \(\textstyle u = \cos(x) \) in the integral, rewrite it as \(\textstyle \int \frac{\tan(x)}{\sqrt{9-u^2}} \cdot (-\frac{du} {\tan(x)}) \). Simplifying this, we get \(\textstyle -\int \frac{1}{\sqrt{9-u^2}} du \).

Step 18: Recognize and Integrate

Recognize that \(\textstyle -\int \frac{1}{\sqrt{9-u^2}} du \) is a standard integral of the form \(\textstyle \int \frac{1}{\sqrt{a^2-u^2}} du \), whose integral is \(\textstyle \sin^{-1}(\frac{u}{a}) \). Here, \(\textstyle a = 3 \). Thus, \(\textstyle - \sin^{-1}(\frac{u}{3}) + C \).

Step 19: Substitute back to x

Finally, substituting back for \(\textstyle u = \cos(x) \), we have \(\textstyle - \sin^{-1}(\frac{\cos(x)}{3}) + C\ = - \arcsin (\frac{\textstyle cos(x)}{\textstyle 3}) + \ C + C \).

Step 20: Confirmation using CAS

Use a Computer Algebra System (CAS) to evaluate the original integral and confirm that \(\textstyle \int \frac{\tan(x)}{\sqrt{9-\cos^2(x)}}dx = - \sin^{-1}(\frac{\cos(x)}{3}) + C \).

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics focused on the concept of integration. Integration is used to find areas, volumes, central points, and many useful things. Essentially, it deals with the accumulation of quantities and the areas under and between curves.
In the exercises provided, we use integration to solve problems involving radical functions. Integral calculus involves a variety of techniques to find these integrals, including basic rules like the power rule and more complex methods like substitution and integration by parts.
Understanding the fundamentals of integral calculus is essential for tackling more complex integrals involving functions like radicals. In each example provided, the solution involves rewriting the integrand (the function being integrated) in a more manageable form, allowing us to apply basic integration rules.

U-substitution

U-substitution, also known as variable substitution, is a technique used to simplify integrals. The idea is to transform a complicated integral into a simpler one by changing the variable of integration.
Take the integral \(\textstyle \textstyle \int \frac{1}{x \sqrt{9x^2+25}}dx\), for example. Here, we use the substitution \(u=9x^2+25\).
When using u-substitution:

• First, identify a substitution that will simplify the integral. For instance, \(\textstyle u=9x^2+25\).
• Derive the differential du related to dx. In this case, \(\textstyle du = 18x \,dx\) or \( dx=\frac{du}{18x}\).
• Rewrite the integral in terms of u. After substitution, this becomes \(\frac{1}{18}\int\frac{1}{u^{\frac{1}{2}}} du\).
• Integrate with respect to u, then substitute back to the original variable x to find the final answer.

In essence, u-substitution turns a complex integral into a simpler form, making it easier to find the solution.

Radical Functions

Radical functions contain roots, such as square roots or cube roots, of polynomials. Integrals involving radicals can often be challenging and require special techniques like u-substitution.
For example, consider \(\textstyle\int e^{x}\sqrt{4+e^{2x}}dx\textstyle\). By substituting \(u = 4 + e^{2x}\), we simplify the integral under the square root into a more recognizable form for integration.
When dealing with radical functions:

• Identify the inner function inside the radical that might be useful for substitution.
• Perform the substitution and rewrite the integrand.
• Simplify the integrand, making it easier to integrate.
• Substitute back to the original variable after integration.

Radical functions often appear in integrals that represent physical quantities involving area and volume in mathematics and physics.

Integration Techniques

Several techniques can be applied to find the integral of complex functions. Knowing when to apply each method is crucial.
Some common integration techniques are:

• U-substitution: As seen in the initial examples, suitable when the integrand contains a composite function.
• Integration by parts: Useful when the integrand is a product of functions. This technique is based on the product rule for differentiation.
• Trigonometric integrals: Useful for integrals involving trigonometric functions, often involving identities or substitutions using trigonometric functions.
• Partial fractions: Useful for integrating rational functions by decomposing them into simpler fractions.

In the provided examples, we primarily used u-substitution to simplify the radical functions. Each technique requires practice and familiarity to identify its best use. Mastering these methods is the key to solving a wide range of integral problems.