Problem 35 - AP Calculus with 8 Practice Tests 15th

Question

\(\int \frac{d x}{\sin ^{2} 2 x}=\) (A) \(\frac{1}{2} \csc 2 x \cot 2 x+C\) (B) \(-\frac{1}{2} \cot 2 x+C\) (C) \(-\cot x+C\) (D) \(-\csc 2 x+C\)

Step-by-Step Solution

Answer

The answer is option (B): \(-\frac{1}{2} \cot 2x + C\).

Step 1: Rewrite the Integrand

The integrand \( \frac{1}{\sin^2 2x} \) is equivalent to \( \csc^2 2x \). Thus, the integral can be rewritten as \( \int \csc^2 2x \, dx \).

Step 2: Use a Trigonometric Identity

Recall that the derivative of \( \cot u \) is \(-\csc^2 u\). Therefore, the integral \( \int \csc^2 2x \, dx \) can be seen as the derivative of \(-\cot 2x\).

Step 3: Apply the Chain Rule

Since we have an inner function \( u = 2x \), we must adjust for the derivative with respect to \( x \). Applying the chain rule gives us \( \int \csc^2 2x \, dx = -\frac{1}{2} \cot 2x + C \). The factor \( \frac{1}{2} \) accounts for the derivative of \( 2x \) which is \( 2 \).

Step 4: Verify the Answer

Check the answer against the given options. The correct answer after considering the integration procedure and chain rule is (B) \(-\frac{1}{2} \cot 2x + C \).

Key Concepts

Integration Techniques

When dealing with integrals like \( \int \frac{1}{\sin ^{2} 2 x} \, dx \), selecting the right integration technique is essential. Here, we are working with a trigonometric integral, which often involves trigonometric identities to simplify the expression.
For this problem, rewriting the integrand \( \frac{1}{\sin^2 2x} \) as \( \csc^2 2x \) is a crucial first step. This substitution allows us to recognize the function as one for which we know the antiderivative.
Typically, in trigonometric integrals, we also look for opportunities to apply other techniques such as substitution or integration by parts. However, in this instance, rewriting the function using a trigonometric identity serves as the main technique.

Trigonometric Identities

Trigonometric identities are invaluable tools for solving integrals involving trigonometric functions.
In this case, recognizing that \( \frac{1}{\sin^2 2x} \) is equivalent to \( \csc^2 2x \) simplifies the integration process significantly. The cosecant-squared function, \( \csc^2 u \), is the derivative of \(-\cot u\).
Thus, using the identity that ties \( \csc^2 u \) with its antiderivative, \(-\cot u\), directly leads to a straightforward evaluation of the integral. It is essential to remember trigonometric identities like \( \csc u = \frac{1}{\sin u} \) and how they interact with derivatives, as they are the key to simplifying and solving more complex integrals.

Chain Rule

The chain rule is a fundamental concept when taking derivatives, and it is equally important when working backwards in integrals.
In this exercise, we're integrating a function of \( 2x \). Recognizing \( u = 2x \) as an inner function is crucial for correct integration. The chain rule states that when differentiating a composite function, such as \( \cot 2x \), you must multiply the derivative of the outer function by the derivative of the inner function.
Similarly, when integrating, we account for the inner function's derivative. Here, the derivative of \( 2x \) is \( 2 \), so we adjust by multiplying the resulting antiderivative \(-\cot 2x\) by \( \frac{1}{2} \). This adjustment ensures that our integration accurately reflects the original structure of the composite function.