Question

Prove that: \( \int_0^{\pi} \log(1 + \cos x) dx = -\pi \log_e 2 \)

Hint:

This problem relies on a famous result: \( \int_0^{\pi/2} \log(\sin x) dx = -\frac{\pi}{2} \log 2 \). Memorizing this and other standard definite integral results can save a lot of time and effort in exams.

Answer 1

Step 1: Understanding the Concept:
This definite integral can be evaluated using the properties of definite integrals and trigonometric identities. The "King's property" \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \) is a key tool for such problems.
Step 2: Key Formula or Approach:
1. Let \( I = \int_0^{\pi} \log(1 + \cos x) dx \).
2. Apply the property \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \) with \( a = \pi \).
3. Add the two expressions for I.
4. Use trigonometric and logarithm properties to simplify the resulting integral.
5. Use the property \( \int_0^{2a} f(x) dx = 2\int_0^a f(x) dx \) if \( f(2a-x) = f(x) \).
Step 3: Detailed Explanation or Calculation:
Let \( I = \int_0^{\pi} \log(1 + \cos x) dx . (1) \)
Using the property \( \int_0^\pi f(x)dx = \int_0^\pi f(\pi-x)dx \):
\[ I = \int_0^{\pi} \log(1 + \cos(\pi - x)) dx \] Since \( \cos(\pi - x) = -\cos x \):
\[ I = \int_0^{\pi} \log(1 - \cos x) dx . (2) \] Adding equations (1) and (2):
\[ 2I = \int_0^{\pi} [\log(1 + \cos x) + \log(1 - \cos x)] dx \] Using the logarithm property \( \log A + \log B = \log(AB) \):
\[ 2I = \int_0^{\pi} \log((1 + \cos x)(1 - \cos x)) dx \] \[ 2I = \int_0^{\pi} \log(1 - \cos^2 x) dx \] Using the identity \( \sin^2 x + \cos^2 x = 1 \):
\[ 2I = \int_0^{\pi} \log(\sin^2 x) dx = \int_0^{\pi} 2\log(\sin x) dx \] \[ I = \int_0^{\pi} \log(\sin x) dx \] Let's check if the integrand \( f(x) = \log(\sin x) \) is symmetric about \( \pi/2 \). \( f(\pi-x) = \log(\sin(\pi-x)) = \log(\sin x) = f(x) \).
So we can use the property \( \int_0^{2a} f(x) dx = 2\int_0^a f(x) dx \). \[ I = 2\int_0^{\pi/2} \log(\sin x) dx \] There is a standard result for this integral: \( \int_0^{\pi/2} \log(\sin x) dx = -\frac{\pi}{2} \log 2 \).
Let's prove this result. Let \( I_1 = \int_0^{\pi/2} \log(\sin x) dx \). Using \( \int_0^a f(x)dx = \int_0^a f(a-x)dx \), \( I_1 = \int_0^{\pi/2} \log(\sin(\pi/2 - x))dx = \int_0^{\pi/2} \log(\cos x) dx \). Adding these two forms of \( I_1 \): \[ 2I_1 = \int_0^{\pi/2} (\log(\sin x) + \log(\cos x)) dx = \int_0^{\pi/2} \log(\sin x \cos x) dx \] \[ 2I_1 = \int_0^{\pi/2} \log\left(\frac{2\sin x \cos x}{2}\right) dx = \int_0^{\pi/2} \log\left(\frac{\sin 2x}{2}\right) dx \] \[ 2I_1 = \int_0^{\pi/2} (\log(\sin 2x) - \log 2) dx = \int_0^{\pi/2} \log(\sin 2x) dx - \int_0^{\pi/2} \log 2 dx \] Let \( t=2x, dt=2dx \). Limits: \( x=0 \to t=0, x=\pi/2 \to t=\pi \). \[ \int_0^{\pi/2} \log(\sin 2x) dx = \frac{1}{2}\int_0^{\pi} \log(\sin t) dt = \frac{1}{2}(2\int_0^{\pi/2} \log(\sin t) dt) = \int_0^{\pi/2} \log(\sin t) dt = I_1 \] So, \( 2I_1 = I_1 - \frac{\pi}{2} \log 2 \). This gives \( I_1 = -\frac{\pi}{2} \log 2 \).
Now substitute this result back into our main problem: \[ I = 2 I_1 = 2 \left(-\frac{\pi}{2} \log 2\right) = -\pi \log 2 \] Step 4: Final Answer:
Hence, it is proved that \( \int_0^{\pi} \log(1 + \cos x) dx = -\pi \log_e 2 \).