Problem 73 - AP Calculus with 8 Practice Tests 15th

Question

\(\int(\tan \theta-1)^{2} d \theta=\) (A) \(\sec \theta+\theta+2 \ln |\cos \theta|+C\) (B) \(\tan \theta+2 \ln |\cos \theta|+C\) (C) \(\tan \theta-2 \sec ^{2} \theta+C\) (D) \(\tan \theta-2 \ln |\cos \theta|+C\)

Step-by-Step Solution

Answer

The answer is (B) \(\tan \theta + 2 \ln |\cos \theta| + C\).

Step 1: Expand the Integrand

First, expand the expression \((\tan \theta - 1)^2\) using the identity \((a - b)^2 = a^2 - 2ab + b^2\). Thus, we have:\[(\tan \theta - 1)^2 = \tan^2 \theta - 2\tan \theta + 1\].

Step 2: Apply Trigonometric Identity

Recall the trigonometric identity \(\tan^2 \theta = \sec^2 \theta - 1\). Substitute this into the expanded form from Step 1:\[\tan^2 \theta - 2\tan \theta + 1 = (\sec^2 \theta - 1) - 2\tan \theta + 1\].Simplify this to:\[\sec^2 \theta - 2\tan \theta\].

Step 3: Integrate Separately

Now integrate each term separately:1. \(\int \sec^2 \theta \, d\theta = \tan \theta + C_1\).2. \(\int -2\tan \theta \, d\theta = -2 \int \tan \theta \, d\theta = -2(-\ln |\cos \theta|) + C_2 = 2\ln |\cos \theta| + C_2\).

Step 4: Combine the Results

Combine the results from Step 3:\[\tan \theta + 2\ln |\cos \theta| + C\].Where \(C = C_1 + C_2\) is the constant of integration.

Key Concepts

Trigonometric Identities

Trigonometric identities are a set of equalities involving trigonometric functions that hold true for any value of the involved angle. In this exercise, we heavily rely on two important identities:

• The square of the tangent function: \( an^2 \theta = \sec^2 \theta - 1\).
• The expression for the secant function squared: \( an^2 \theta = \sec^2 \theta - 1\) simplifies expressions involving \(\tan \theta\).

The first identity allows us to replace \(\tan^2 \theta\) with \(\sec^2 \theta - 1\), which simplifies our integration process. This identity is extremely handy when solving integrals involving \(\tan \theta\). By applying these identities during problem-solving, calculations become more straightforward and easily manageable. Always look out for opportunities to replace trigonometric expressions with these identities to simplify complex problems. Practice recognizing these patterns to improve both speed and accuracy in mathematics.

Integration Techniques

Integration techniques are methods used to find the integral of functions that would otherwise be difficult to integrate directly. In this exercise, we consider the integral of \(\sec^2 \theta\) and \(-2\tan \theta\) separately. Let's look into some key techniques applied here:

• Separate Integration: Breaking down complicated expressions into simpler parts often simplifies integration. Here, separating \(\int(\sec^2 \theta - 2\tan \theta)\, d\theta\) into \(\int \sec^2 \theta \, d\theta\) and \(\int -2\tan \theta \, d\theta\) helps in managing computations piecemeal.
• Use of Known Integrals: Recognizing the integral of known trigonometric functions such as \(\int \sec^2 \theta \, d\theta = \tan \theta\) and \(\int \tan \theta \, d\theta = -\ln |\cos \theta|\), speeds up the process. Familiarity with these fundamental integrals is extremely useful.

Understanding the application of these techniques enhances problem-solving skills and helps in tackling diverse mathematics challenges.

Definite and Indefinite Integrals

Integrals can be classified into two types: definite and indefinite integrals. In this exercise, we deal with an indefinite integral, which finds an antiderivative of the function without specific bounds.

• Indefinite Integrals: Represented by the symbol \(\int\), they include a constant of integration \(C\) since integration is the reverse process of differentiation. The final solution includes this arbitrary constant.

When computing indefinite integrals:

• Focus on the antiderivative of the function.
• After integrating, ensure the result includes the constant \(C\).

This exercise highlights the importance of correctly identifying the integration's type, as different approaches are needed for definite versus indefinite integrals. Start familiarizing with both types early on, as integrals form the foundation for advanced calculus topics.