Question

Prove that \( \int_{0}^{\pi/2} \sqrt{\frac{1+\cos 4x}{2}} \, dx = 1 \).

Hint:

When simplifying \( \sqrt{f(x)^2} \), always remember to write it as \( |f(x)| \). Forgetting the absolute value is a common error and can lead to incorrect results if \( f(x) \) is negative over part of the integration interval.

Answer 1

Step 1: Understanding the Concept:
This problem requires evaluating a definite integral. The integrand involves a square root of a trigonometric function, which can be simplified using a half-angle or double-angle identity.
Step 2: Key Formula or Approach:
We will use the trigonometric identity for the cosine of a double angle: \[ \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \] By letting \( \theta = 2x \), this identity becomes: \[ \cos^2(2x) = \frac{1 + \cos 4x}{2} \] Step 3: Detailed Explanation:
Let the integral be I. \[ I = \int_{0}^{\pi/2} \sqrt{\frac{1+\cos 4x}{2}} \, dx \] Using the identity from Step 2, we can simplify the expression under the square root: \[ I = \int_{0}^{\pi/2} \sqrt{\cos^2(2x)} \, dx \] \[ I = \int_{0}^{\pi/2} |\cos(2x)| \, dx \] The presence of the absolute value means we must consider the sign of \( \cos(2x) \) over the interval \( [0, \pi/2] \). For \( 0 \leq x \leq \pi/4 \), we have \( 0 \leq 2x \leq \pi/2 \), where \( \cos(2x) \geq 0 \).
For \( \pi/4<x \leq \pi/2 \), we have \( \pi/2<2x \leq \pi \), where \( \cos(2x) \leq 0 \).
Therefore, we must split the integral at \( x = \pi/4 \): \[ I = \int_{0}^{\pi/4} \cos(2x) \, dx + \int_{\pi/4}^{\pi/2} (-\cos(2x)) \, dx \] Now, we evaluate each integral: \[ \int \cos(2x) \, dx = \frac{\sin(2x)}{2} \] For the first part: \[ \left[ \frac{\sin(2x)}{2} \right]_{0}^{\pi/4} = \frac{\sin(2 \cdot \pi/4)}{2} - \frac{\sin(0)}{2} = \frac{\sin(\pi/2)}{2} - 0 = \frac{1}{2} \] For the second part: \[ \left[ -\frac{\sin(2x)}{2} \right]_{\pi/4}^{\pi/2} = \left(-\frac{\sin(2 \cdot \pi/2)}{2}\right) - \left(-\frac{\sin(2 \cdot \pi/4)}{2}\right) = -\frac{\sin(\pi)}{2} + \frac{\sin(\pi/2)}{2} = 0 + \frac{1}{2} = \frac{1}{2} \] Adding the results from both parts: \[ I = \frac{1}{2} + \frac{1}{2} = 1 \] Step 4: Final Answer:
We have shown that the value of the definite integral is 1.