Question

Prove that \(\int_0^{\pi/2} \log(\cos x) \, dx = -\frac{\pi}{2} \log 2\).

Hint:

This is a standard result in definite integration and is very useful to remember. Both \(\int_0^{\pi/2} \log(\cos x) dx\) and \(\int_0^{\pi/2} \log(\sin x) dx\) are equal to \(-\frac{\pi}{2}\log 2\). The property \(\int_0^a f(x) dx = \int_0^a f(a-x) dx\) is extremely powerful for integrals with limits 0 to a.

Answer 1

Step 1: Understanding the Concept:
This is a classic problem in definite integrals. It requires the use of the properties of definite integrals, particularly the property \(\int_0^a f(x) dx = \int_0^a f(a-x) dx\). This allows us to create a second related integral which, when combined with the first, leads to a simpler expression that can be evaluated.
Step 2: Key Formula or Approach:
1. Let the given integral be \(I\). So, \(I = \int_0^{\pi/2} \log(\cos x) dx\).
2. Apply the property \(\int_0^a f(x) dx = \int_0^a f(a-x) dx\) with \(a = \pi/2\).
3. This will create a second equation for \(I\) in terms of \(\log(\sin x)\).
4. Add the two equations for \(I\). This will combine the logarithms into \(\log(\sin x \cos x)\).
5. Use the identity \(\sin(2x) = 2\sin x \cos x\) to simplify the logarithm.
6. Split the resulting integral and solve for \(I\).
Step 3: Detailed Explanation:
Let \(I = \int_0^{\pi/2} \log(\cos x) dx \quad \cdots (1)\).
Using the property \(\int_0^a f(x) dx = \int_0^a f(a-x) dx\): \[ I = \int_0^{\pi/2} \log(\cos(\pi/2 - x)) dx \] Since \(\cos(\pi/2 - x) = \sin x\), we have: \[ I = \int_0^{\pi/2} \log(\sin x) dx \quad \cdots (2) \] Now, add equations (1) and (2): \[ 2I = \int_0^{\pi/2} \log(\cos x) dx + \int_0^{\pi/2} \log(\sin x) dx \] \[ 2I = \int_0^{\pi/2} (\log(\cos x) + \log(\sin x)) dx \] Using the logarithm property \(\log A + \log B = \log(AB)\): \[ 2I = \int_0^{\pi/2} \log(\sin x \cos x) dx \] To use the double angle formula, multiply and divide by 2 inside the logarithm: \[ 2I = \int_0^{\pi/2} \log\left(\frac{2\sin x \cos x}{2}\right) dx = \int_0^{\pi/2} \log\left(\frac{\sin(2x)}{2}\right) dx \] Using the property \(\log(A/B) = \log A - \log B\): \[ 2I = \int_0^{\pi/2} \log(\sin(2x)) dx - \int_0^{\pi/2} \log 2 dx \] Let's evaluate the second integral first: \[ \int_0^{\pi/2} \log 2 dx = \log 2 [x]_0^{\pi/2} = \frac{\pi}{2}\log 2 \] Now consider the first integral, \(\int_0^{\pi/2} \log(\sin(2x)) dx\). Let \(t = 2x\), so \(dt = 2dx\) or \(dx = dt/2\). The limits of integration change from \(x=0 \to t=0\) and \(x=\pi/2 \to t=\pi\). \[ \int_0^{\pi/2} \log(\sin(2x)) dx = \int_0^{\pi} \log(\sin t) \frac{dt}{2} = \frac{1}{2}\int_0^{\pi} \log(\sin t) dt \] Using the property \(\int_0^{2a} f(x) dx = 2\int_0^a f(x) dx\) if \(f(2a-x)=f(x)\). Here \(2a=\pi\), \(a=\pi/2\). \(\sin(\pi-t) = \sin t\), so the property applies. \[ \frac{1}{2}\int_0^{\pi} \log(\sin t) dt = \frac{1}{2} \cdot 2 \int_0^{\pi/2} \log(\sin t) dt = \int_0^{\pi/2} \log(\sin x) dx = I \] Substituting everything back into the equation for \(2I\): \[ 2I = I - \frac{\pi}{2}\log 2 \] Solving for I: \[ 2I - I = -\frac{\pi}{2}\log 2 \] \[ I = -\frac{\pi}{2}\log 2 \] Step 4: Final Answer:
We have proven that \(\int_0^{\pi/2} \log(\cos x) dx = -\frac{\pi}{2}\log 2\).