Evaluate the integral 0 to pi/2 log cos x dx
00:00 Intro
We apply the property of definite integral to evaluate the integral of log(cos x) as x changes from 0 to π/2. The key steps and properties used are:
• Applying Integral Property (0:06-1:30): The integral is defined as I. The speaker applies the property that ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a-x) dx, replacing 'x' with 'π/2 - x'. This transforms the integral into ∫₀^(π/2) log(sin x) dx, which is still equal to I.
• Adding the Integrals (1:48-3:29): The two expressions for I (log(cos x) and log(sin x)) are added together, resulting in 2I = ∫₀^(π/2) (log(sin x) + log(cos x)) dx. Using the logarithm property log a + log b = log (ab), this simplifies to 2I = ∫₀^(π/2) log(sin x cos x) dx.
• Trigonometric Identity and Logarithm Property (3:43-4:52): The expression inside the logarithm is manipulated using the identity sin 2x = 2 sin x cos x, so sin x cos x = (sin 2x)/2. This leads to 2I = ∫₀^(π/2) log((sin 2x)/2) dx. Applying the logarithm property log(a/b) = log a - log b, the integral becomes 2I = ∫₀^(π/2) (log(sin 2x) - log 2) dx.
• Substitution and Limit Change (5:56-7:20): A substitution is made, letting t = 2x. This changes the limits of integration from 0 to π (as x goes from 0 to π/2, t goes from 0 to π) and dx becomes (1/2) dt. The integral for 2I is now (1/2) ∫₀^π log(sin t) dt - (π/2) log 2.
• Applying Another Integral Property (7:50-9:19): The property ∫₀^(2a) f(x) dx = 2 ∫₀ᵃ f(x) dx (if f(2a-x) = f(x)) is used for the log(sin t) integral. Since sin(π - t) = sin t, the integral (1/2) ∫₀^π log(sin t) dt becomes (1/2) * 2 ∫₀^(π/2) log(sin t) dt, which simplifies to ∫₀^(π/2) log(sin t) dt.
• Final Solution (9:35-10:41): This simplified integral is recognized as the original 'I' (since 't' is a dummy variable). So the equation becomes 2I = I - (π/2) log 2. Solving for I, we get I = -(π/2) log 2.