Problem 26 - AP Calculus with 8 Practice Tests 15th

Question

\(\int \cos ^{2} 2 x d x=\) (A) \(\frac{x}{2}+\frac{\sin 4 x}{8}+C\) (B) \(\frac{x}{2}-\frac{\sin 4 x}{8}+C\) (C) \(\frac{x}{4}+\frac{\sin 4 x}{4}+C\) (D) \(\frac{x}{4}+\frac{\sin 4 x}{16}+C\)

Step-by-Step Solution

Answer

The correct answer is (A).

Step 1: Use the Power-Reducing Identity

First, recognize that the integrand \( \cos^2 2x \) needs to be simplified. Use the power-reducing identity for cosine, which is \( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \). Applying this to \( \cos^2 2x \), we get: \[\cos^2 2x = \frac{1 + \cos 4x}{2}\]

Step 2: Rewrite the Integral Using the Identity

Substitute \( \cos^2 2x \) from Step 1 into the integral: \[\int \cos^2 2x \, dx = \int \frac{1 + \cos 4x}{2} \, dx\] This can be split into two separate integrals: \[\int \frac{1}{2} \, dx + \int \frac{\cos 4x}{2} \, dx\]

Step 3: Evaluate the First Integral

The first integral is straightforward: \[\int \frac{1}{2} \, dx = \frac{x}{2}\]

Step 4: Evaluate the Second Integral

For the second integral, \( \int \frac{\cos 4x}{2} \, dx \), recognize this as a simple u-substitution problem, where \( u = 4x \), thus \( du = 4 \, dx \) or \( dx = \frac{1}{4} du \). The integral becomes: \[\frac{1}{2} \int \cos 4x \, dx = \frac{1}{2} \cdot \frac{1}{4} \int \cos u \, du\] \[= \frac{1}{8} \int \cos u \, du = \frac{1}{8} \sin u + C = \frac{1}{8} \sin 4x + C\]

Step 5: Combine the Results

Combine the results from Steps 3 and 4: \[\frac{x}{2} + \frac{1}{8} \sin 4x + C\] Simplifying, the solution to the integral is: \( \frac{x}{2} + \frac{\sin 4x}{8} + C \).

Step 6: Identify the Correct Multiple Choice Answer

Compare our result with the given choices. Our solution \( \frac{x}{2} + \frac{\sin 4x}{8} + C \) matches option (A).

Key Concepts

Power-Reducing Identity

The power-reducing identity is a useful tool in integral calculus, particularly when dealing with trigonometric functions raised to a power, like \(\cos^2 2x\). This identity helps in simplifying expressions, making it easier to perform integration. For cosine, the power-reducing identity is

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

This formula transforms a square of a cosine function into a linear combination, involving a constant and a cosine of double the angle.
So, for \(\cos^2 2x\), applying this identity gives us \(\frac{1 + \cos 4x}{2}\).
This step simplifies the integral significantly, paving the way for easier integration using basic calculus techniques.

u-substitution

u-substitution is a method used to evaluate integrals, where a substitution is made to simplify the integral into an easily solvable form.

• It's especially useful when dealing with composite functions.
• The idea is to choose a substitution that turns the integral into a simpler function of a new variable, "u".

In the context of \(\int \frac{\cos 4x}{2} \, dx\), we use \(u = 4x\), implying \(du = 4 \, dx\). Therefore, \(dx = \frac{1}{4} \, du\).
This substitution transforms the integral into \(\frac{1}{8} \int \cos u \, du\), which is more straightforward to solve.
After substituting back, we arrive at the solution of \(\frac{1}{8} \sin 4x + C\). This method is vital for solving integrals that are not standard or directly integrable.

Trigonometric Integrals

Trigonometric integrals involve integrating expressions containing trigonometric functions. Working with these can often be tricky because they can involve powers and products of trigonometric functions.

• Common strategies include using trigonometric identities, like the power-reducing identity, and techniques such as substitution.
• Trigonometric integrals often arise in computing areas under curves, physics problems, and other applied mathematics contexts.

In the problem given, the integral \(\int \cos^2 2x \, dx\) is tackled by simplifying using the power-reducing identity, then using u-substitution on the resultant integral part.
Both methods combined simplify the process, breaking down a complex integral into manageable parts. Understanding these concepts can make solving trigonometric integrals more intuitive and straightforward.