Question

Prove that \(\int_0^\pi \sqrt{\frac{1+\cos 2x}{2}} \, dx = 2\).

Hint:

A common mistake is to simplify \(\sqrt{\cos^2 x}\) to just \(\cos x\). Always remember that the square root of a squared quantity is its absolute value, i.e., \(\sqrt{u^2} = |u|\). This often requires splitting the integral into multiple parts based on the sign of the function inside the absolute value.

Answer 1

Step 1: Understanding the Concept:
This problem involves evaluating a definite integral. The key to solving it is to simplify the integrand using a trigonometric identity before performing the integration. We must also be careful with the square root, as \(\sqrt{u^2} = |u|\).
Step 2: Key Formula or Approach:
We will use the half-angle trigonometric identity: \[ \cos^2 x = \frac{1 + \cos 2x}{2} \] After substitution, the integral will involve \(|\cos x|\), which needs to be handled by splitting the integral over intervals where \(\cos x\) is positive and negative.
Step 3: Detailed Explanation:
Let the integral be I. \[ I = \int_0^\pi \sqrt{\frac{1+\cos 2x}{2}} dx \] Using the identity \(\cos^2 x = \frac{1+\cos 2x}{2}\), we get: \[ I = \int_0^\pi \sqrt{\cos^2 x} \, dx \] \[ I = \int_0^\pi |\cos x| \, dx \] Now, we need to consider the sign of \(\cos x\) in the interval \([0, \pi]\).
For \(x \in [0, \frac{\pi}{2}]\), \(\cos x \geq 0\), so \(|\cos x| = \cos x\).
For \(x \in (\frac{\pi}{2}, \pi]\), \(\cos x \leq 0\), so \(|\cos x| = -\cos x\).
We split the integral at \(x = \frac{\pi}{2}\): \[ I = \int_0^{\pi/2} \cos x \, dx + \int_{\pi/2}^\pi (-\cos x) \, dx \] Now, we evaluate each part: \[ \int_0^{\pi/2} \cos x \, dx = [\sin x]_0^{\pi/2} = \sin(\frac{\pi}{2}) - \sin(0) = 1 - 0 = 1 \] \[ \int_{\pi/2}^\pi (-\cos x) \, dx = [-\sin x]_{\pi/2}^\pi = (-\sin(\pi)) - (-\sin(\frac{\pi}{2})) = 0 - (-1) = 1 \] Adding the results of the two parts: \[ I = 1 + 1 = 2 \] Step 4: Final Answer:
We have shown that the value of the integral is 2. Hence proved.