Problem 38 - AP Calculus with 8 Practice Tests 15th

Question

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

Step-by-Step Solution

Answer

(B) \(-\ln |1 - \cos 2 t| + C\).

Step 1: Simplify the Integrand

The integrand is \( \frac{\sin 2t}{1 - \cos 2t} \). Using trigonometric identities, we know that \( \sin 2t = 2\sin t \cos t \) and \( \cos 2t = 1 - 2\sin^2 t \). Therefore, \( 1 - \cos 2t = 2\sin^2 t \). Substituting these, we get \( \frac{\sin 2t}{1 - \cos 2t} = \frac{2\sin t \cos t}{2\sin^2 t} = \frac{\cos t}{\sin t} = \cot t \).

Step 2: Integral of cotangent

The integral \( \int \cot t \, dt\) is equal to \( \ln |\sin t| + C \). This uses the standard integral \( \int \cot u \, du = \ln |\sin u| + C \).

Step 3: Substitute back using double angle identity

Since \( t \) is replaced by \( 2t \), change the variable back. Let us note that \( \sin 2t = 1 - \cos 2t \) and thus the substitution \( u = \sin 2t \) gives \( |\sin t| = |1 - \cos 2t| \). So \( \ln |\sin t| + C \) becomes \( \ln |1 - \cos 2t| + C \).

Step 4: Check the options

The negative sign must be considered: From differentiation, we know that the derivative of the integral must be the original integrand. Therefore, the correct option must account for this step. The correct form is \(-\ln|1 - \cos 2t| + C \).

Step 5: Final Answer

After comparing with the provided options, the integral \( \int \frac{\sin 2t}{1 - \cos 2t} \, dt \) corresponds to choice (B): \(-\ln |1 - \cos 2t| + C \).

Key Concepts

Integration Techniques

Integral calculus often involves finding the antiderivative or integral of a function. To solve complex integrals, we use **integration techniques**. These methods help simplify the integrand or change the variable to make integration more manageable.
**Common Integration Techniques Include**:

• Substitution Method: Used when the integrand is a composite function. It simplifies the integral by changing the variable.
• Integration by Parts: Useful when the integrand is a product of two functions.
• Partial Fraction Decomposition: Breaks down rational functions into simpler fractions that are easier to integrate.

In this problem, the substitution method is used alongside trigonometric identities to simplify the integrand. Understanding which technique to apply is often key to finding the solution.

Trigonometric Identities

**Trigonometric identities** are relationships involving trigonometric functions that hold true for all values of the involved variables. They are immensely powerful in calculus for simplifying expressions.
For the integrand \( \frac{\sin 2t}{1-\cos 2t} \), the following identities were crucial:

• Double Angle Formulas: \( \sin 2t = 2 \sin t \cos t \) and \( \cos 2t = 1 - 2\sin^2 t \)
• Pythagorean Identities: Expresses squares of trigonometric functions in terms of each other.

Understanding these identities, we made the expression simpler by expressing \( 1 - \cos 2t \) using\( \sin^2 t \), eventually simplifying the integrand to a form that is easier to integrate, i.e., \( \frac{\cos t}{\sin t} = \cot t \). This is a neat demonstration of using identities to transform complex expressions.

Integral of Trigonometric Functions

Calculating the **integral of trigonometric functions** is a common task in calculus. Each basic trigonometric function has a standard integral. For cotangent, it is crucial to know the integral:The integral \( \int \cot t \, dt \) is equal to \( \ln |\sin t| + C \).
Here's the thought process:- Recognize the function form: In our case, the simplified integrand is \( \cot t \).- Recall the standard integral for \( \cot \), which is \( \ln |\sin t| + C \).**After Integrating the Simplified Expression:**When solving the integral, remember to re-substitute any variable changes made during the simplification phase. We switched from \( \sin t \) back to the variable related to original expression, \( 1-\cos 2t \), resulting in \( \ln |1-\cos 2t| + C \). To match the provided answer options, the formula was adjusted to \(-\ln |1-\cos 2t| + C \). This highlights the integral's sensitivity to sign, especially when verifying results with given choices.