Problem 22 - AP Calculus with 8 Practice Tests 15th

Question

\(\int \frac{u d u}{\sqrt{4-9 u^{2}}}=\) (A) \(\frac{1}{3} \sin ^{-1} \frac{3 u}{2}+C\) (B) \(-\frac{1}{18} \ln \sqrt{4-9 u^{2}}+C\) (C) \(2 \sqrt{4-9 u^{2}}+C\) (D) \(-\frac{1}{9} \sqrt{4-9 u^{2}}+C\)

Step-by-Step Solution

Answer

The correct answer is (D) \(-\frac{1}{9} \sqrt{4-9 u^{2}}+C\).

Step 1: Identify the Integral Form

The integral \( \int \frac{u \, du}{\sqrt{4-9u^2}} \) hints at a trigonometric substitution, as the denominator is a square root difference of a constant and a variable expression, resembling the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \).

Step 2: Substitute and Simplify

Let \( 9u^2 = 9\sin^2(\theta) \) , thus \( u = \frac{2}{3}\sin(\theta) \) and \( du = \frac{2}{3}\cos(\theta) \, d\theta \). This changes the integral to \( \int \frac{\frac{2}{3}\sin(\theta) \cdot \frac{2}{3}\cos(\theta)\,d\theta}{\sqrt{4-9\left(\frac{2}{3}\sin(\theta)\right)^2}} \).

Step 3: Simplify the Denominator

The denominator simplifies to \( \sqrt{4 - 4\sin^2(\theta)} = \sqrt{4(1-\sin^2(\theta))} = \sqrt{4\cos^2(\theta)} = 2\cos(\theta) \).

Step 4: Evaluate the Integral

Now, substitute back into the integral: \( \int \frac{\frac{4}{9}\sin(\theta)\cos(\theta)\,d\theta}{2\cos(\theta)} = \frac{2}{9} \int \sin(\theta)\,d\theta \). The integral of \( \sin(\theta) \) is \( -\cos(\theta) \), so the integral becomes \( -\frac{2}{9}\cos(\theta) + C \).

Step 5: Convert Back to Original Variable

Recall \( u = \frac{2}{3}\sin(\theta) \), thus \( \sin(\theta) = \frac{3u}{2} \). Therefore, \( \cos(\theta) = \sqrt{1-\sin^2(\theta)} = \sqrt{1-\left(\frac{3u}{2}\right)^2} = \frac{\sqrt{4-9u^2}}{2} \). Substitute back: \( -\frac{2}{9} \cdot \frac{\sqrt{4-9u^2}}{2} = -\frac{1}{9}\sqrt{4-9u^2} + C \).

Step 6: Match Solution to Given Options

Compare this final form with the options. The equivalent choice is \( -\frac{1}{9}\sqrt{4-9u^2} + C \), which is choice (D).

Key Concepts

Integral Calculus

Integral calculus is a significant branch of mathematics that deals with integrals and the area under curves. It's about finding anti-derivatives or the original function given its derivative.
In simpler terms, while differentiation breaks a function into its rate of change, integration helps in assembling it back from its rate of change.When dealing with integrals, especially those involving functions under square roots, trigonometric substitutions can be very helpful. These are techniques used to simplify integrands. In this exercise, we see such application where we are asked to solve an integral of the form \( \int \frac{u \, du}{\sqrt{4-9u^2}} \). Utilizing calculus ideas to simplify complex expressions is crucial in comprehension and problem-solving.

Integration Techniques

Integration techniques involve various strategies to solve integrals, especially those complicated or non-standard forms. Understanding which technique to apply is key:

• Substitution Method: This involves changing variables to simplify the integral.
• Integration by Parts: Useful for products of functions, derived from the product rule of differentiation.
• Partial Fractions: Useful for rational functions, breaking them into simpler parts.
• Trigonometric Substitution: Used when the integral contains square roots or expressions involving \( a^2 \pm u^2 \).

In the original solution given, trigonometric substitution is the chosen technique. This turns a complex expression into something more tractable by leveraging trigonometric identities. For example, the step involves substituting terms like \( u = \frac{2}{3}\sin(\theta) \) to simplify computation. This approach transforms a challenging integral into one easier to calculate.

Trigonometric Functions

Trigonometric functions such as sine and cosine are fundamental in mathematics and are especially useful in integration techniques.They come from right-angled triangles and the unit circle and have equations such as:

• \( \sin^2(\theta) + \cos^2(\theta) = 1 \)
• \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \)

In integration, these functions can simplify the process by transforming complex algebraic expressions. This exercise showcases such a transformation, where trigonometric identities and functions help recreate the integral's variable back from trigonometric substitutions.By rewriting \( u \) in terms of \( \sin(\theta) \), and simplifying the expression using \( \cos(\theta) \), the given integral \( \int \frac{u \, du}{\sqrt{4-9u^2}} \) can be resolved step by step using the inherent relationships within trigonometric functions. Understanding these relationships allows solving integrals that might initially appear daunting.