Problem 400 - Calculus Volume 1

Question

In the following exercises, find each indefinite integral, using appropriate substitutions. $$\int \frac{d x}{25+16 x^{2}}$$

Step-by-Step Solution

Answer

The indefinite integral is \( \frac{1}{20} \tan^{-1} \left( \frac{4x}{5} \right) + C \).

Step 1: Recognize the Form

The given integral \( \int \frac{d x}{25+16 x^{2}} \) is of the form \( \int \frac{d x}{a^{2} + u^{2}} \), which resembles the standard integral formula for the inverse tangent: \( \int \frac{d u}{a^{2} + u^{2}} = \frac{1}{a} \tan^{-1} \left( \frac{u}{a} \right) + C \). In our integral, \( a^2 = 25 \) and \( 16x^2 = u^2 \).

Step 2: Identify and Apply Substitution

Let \( u = 4x \). Then, \( du = 4 \, dx \), or \( dx = \frac{1}{4} \, du \). This substitution transforms \( 25 + 16x^2 \) into \( 25 + u^2 \), simplifying our integral.

Step 3: Substitute and Simplify the Integral

Substituting \( u = 4x \) and \( dx = \frac{1}{4} \, du \) into the integral gives: \( \int \frac{1}{25 + u^2} \cdot \frac{1}{4} \, du \). Simplify this to \( \frac{1}{4} \int \frac{du}{25 + u^2} \).

Step 4: Apply the Formula for Arc Tangent

Using the standard integral formula \( \int \frac{d u}{a^{2} + u^{2}} = \frac{1}{a} \tan^{-1} \left( \frac{u}{a} \right) + C \) with \( a = 5 \) (since \( a^2 = 25 \)), the integral \( \frac{1}{4} \int \frac{du}{25 + u^{2}} \) becomes \( \frac{1}{4} \left( \frac{1}{5} \tan^{-1} \left( \frac{u}{5} \right) \right) + C \).

Step 5: Substitute Back in Terms of x

Replace \( u \, \) back with \( 4x \), transforming the expression into \( \frac{1}{20} \tan^{-1} \left( \frac{4x}{5} \right) + C \), which is the solution of the original integral.

Key Concepts

Integration by Substitution

Integration by substitution is like reverse engineering a complex problem into a simpler one. By using this technique, difficult integrals can be made manageable by transforming the variable being integrated. Consider the example of substituting for a function that makes the integral fall into a known form, like using the trigonometric identity or inverse functions.

• Identify a part of the integral that matches a standard form.
• Choose a substitution that simplifies the expression.
• Swap the variables and adjust the differential accordingly.

With the problem at hand, we replace the original variable, which was less straightforward, with a new one, such that the integral assumes a form we recognize. Here, we used the substitution \( u = 4x \), which simplifies \( 16x^2 \) to \( u^2 \), leading to a more standard integral form.

Inverse Trigonometric Functions

Inverse trigonometric functions such as the inverse tangent help us evaluate integrals that involve specific expressions. These functions provide a bridge from algebraic forms back to angles or circular functions, making them incredibly useful in calculus. In our exercise, the resemblance of our transformed integral to the arctangent formula is clear. We have:\[ \int \frac{du}{25 + u^2} = \frac{1}{5} \tan^{-1} \left( \frac{u}{5} \right) + C \]Understanding that this kind of integral often leads to inverse trigonometric results allows us to solve complex integrals efficiently. Recognizing these patterns is key:

• \( \int \frac{du}{a^2 + u^2} \rightarrow \frac{1}{a} \tan^{-1} \left( \frac{u}{a} \right) + C \)
• These patterns make evaluating integrals much easier.

Thus, once we transformed \( 25 + 16x^2 \) to \( 25 + u^2 \), it fit the arc tangent's form perfectly, simplifying our computation significantly.

Integration Techniques

There are many techniques in integration that can simplify or solve even the toughest integrals. Understanding when and how to apply different methods is crucial for becoming proficient at calculus. Let's consider some of the most commonly used techniques:

• Substitution: Ideal for integrals containing a composite function or when a simple derivative appears.
• Partial Fractions: Useful for rational functions where a polynomial is divided by another.
• Trigonometric Identities: Helpful when integrals involve trigonometric functions needing simplification.

In our specific example, we primarily leaned on substitution to reduce a complex form into one we recognize. This allowed the integral to fit beautifully into the framework of inverse trigonometric functions. Each transformation step brought us closer to a solution that was much simpler than initially encountered. While there are multiple methods available, choosing the correct technique results in clearer solutions and reduces computational hassle, as demonstrated through the example.