Problem 27 - AP Calculus with 8 Practice Tests 15th

Question

\(\int \sin 2 \theta d \theta=\) (A) \(\frac{1}{2} \cos 2 \theta+C\) (B) \(-2 \cos 2 \theta+C\) (C) \(\cos ^{2} \theta+C\) (D) \(-\frac{1}{2} \cos 2 \theta+C\)

Step-by-Step Solution

Answer

The correct option is (D) \(-\frac{1}{2} \cos 2 \theta+C\).

Step 1: Identify the Integral to Solve

We need to solve the integral \( \int \sin 2 \theta \, d\theta \). This integral involves a trigonometric function with a multiple angle.

Step 2: Use Substitution Method

Let's use the substitution method to make the integral simpler. Set \( u = 2\theta \), then \( du = 2 \, d\theta \), or \( d\theta = \frac{1}{2} du \). Substitute these into the integral to get \( \int \sin u \frac{1}{2} du \).

Step 3: Solve the Integral

Next, solve \( \int \frac{1}{2} \sin u \, du \). Since the integral of \( \sin u \) is \( -\cos u \), we have: \[ \int \frac{1}{2} \sin u \, du = -\frac{1}{2} \cos u + C \]

Step 4: Substitute Back Original Variable

Replace \( u \) with \( 2\theta \) to return to the original variable. So we have: \[ -\frac{1}{2} \cos(2\theta) + C \]

Step 5: Identify the Correct Answer Choice

Compare the result \( -\frac{1}{2} \cos(2\theta) + C \) to the provided options. The correct option is (D) \(-\frac{1}{2} \cos 2 \theta+C\).

Key Concepts

Substitution Method

One of the foundational integration techniques in calculus is the substitution method. This method is particularly helpful when dealing with trigonometric integrals like \( \int \sin 2 \theta \, d\theta \). The main idea is to simplify the integral by changing variables.

We start by setting a new variable \( u \) equal to the inner function of a composite function. In our exercise, we set \( u = 2\theta \). This changes our integral with respect to \( \theta \) to an integral in terms of \( u \).

Next, we differentiate \( u \) to find \( du \), giving us \( du = 2 \, d\theta \). Thus, \( d\theta \) can be expressed as \( \frac{1}{2} du \). Substituting these into our integral, we re-write \( \int \sin 2 \theta \, d\theta \) as \( \int \sin u \frac{1}{2} du \).

By substituting, we've transformed a potentially complex integral into a much simpler form, making it possible to integrate easily.

Multiple Angle Formulas

When integrating trigonometric functions, you'll often encounter expressions involving multiple angles like \( \sin 2\theta \). Recognizing these can be vital because multiple angle formulas enable us to simplify these expressions.

The formula for \( \sin 2\theta \) is used frequently: \( \sin 2\theta = 2 \sin \theta \cos \theta \). However, in this exercise, rather than using this identity to simplify directly, we used substitution to handle \( \sin 2\theta \) as a single variable transformation.

Understanding multiple angle formulas is still important because they can offer alternative solution paths in more complex transformations or when combining with other trigonometric identities.

This knowledge enhances problem-solving skills by opening up different approaches to tackle integrals involving trigonometric functions.

Calculus Integration Techniques

Calculus is enriched with various integration techniques, and mastering them allows for solving diverse mathematical problems. In our example problem, the substitution method was showcased, but other techniques like parts and partial fractions can also be useful.

The substitution method simplifies the process when dealing with composite functions or integrals involving trigonometric identities. Another common technique is integration by parts, useful for products of functions.

• Integration by parts is based on the product rule of differentiation.
• It is suitable when the integrand is a product of two distinct functions.

Partial fraction decomposition is a technique often used when the integral involves rational functions. It breaks down a more complex rational expression into simpler fractions that can be integrated separately.

Broadening your understanding of various calculus integration techniques provides you with a toolkit to tackle different types of integrals efficiently.