Question

If \(\int_0^{\pi/2} \log(\cos x) \, dx = \frac{\pi}{2} \log\left(\frac{1}{2}\right)\), then find \(\int_0^{\pi/2} \log(\sec x) \, dx\).

Answer 1

To find ∫₀^(π/2) log(sec(x)) dx, we can use the properties of logarithms and the trigonometric identity sec(x) = 1/cos(x).

First, rewrite the integral using the identity:

∫₀^(π/2) log(sec(x)) dx = ∫₀^(π/2) log(1/cos(x)) dx

Next, use the property of logarithms:

∫₀^(π/2) log(sec(x)) dx = ∫₀^(π/2) (-log(cos(x))) dx

Now, we can substitute the given value of the integral ∫₀^(π/2) log(cos(x)) dx:

∫₀^(π/2) (-log(cos(x))) dx = π/2 * log(1/2)

Multiplying both sides by -1:

-∫₀^(π/2) log(cos(x)) dx = -π/2 * log(1/2)

Finally, we can substitute back to the original integral:

∫₀^(π/2) log(sec(x)) dx = -π/2 * log(1/2)

Since the left side is the integral we want to evaluate, we can rewrite the equation:

∫₀^(π/2) log(sec(x)) dx = π/2 * log(2)

Therefore, the value of ∫₀^(π/2) log(sec(x)) dx is π/2 * log(2)

Answer 2

Given:\(\int_0^{\pi/2} \log(\cos x) \, dx = \frac{\pi}{2} \log\left(\frac{1}{2}\right)\)

We know that \(\log(\sec x) = -\log(\cos x)\) because \(\sec x = \frac{1}{\cos x}\). Therefore,

\(\int_0^{\pi/2} \log(\sec x) \, dx = \int_0^{\pi/2} -\log(\cos x) \, dx = -\int_0^{\pi/2} \log(\cos x) \, dx\)

Using the given value for \(\int_0^{\pi/2} \log(\cos x) \, dx\), we substitute it into the expression:

\(-\int_0^{\pi/2} \log(\cos x) \, dx = -\left( \frac{\pi}{2} \log\left(\frac{1}{2}\right) \right)\)

\(-\left( \frac{\pi}{2} \log\left(\frac{1}{2}\right) \right) = \frac{\pi}{2} \log(2)\)

So, the answer is: \(\frac{\pi}{2} \log(2)\)