Evaluate: ∫^π/₂ log(sin x) dx

To evaluate the integral ∫ (π/2) log(sin x) dx, we can use a clever technique involving symmetry and properties of definite integrals.

Step 1: Understanding the Integral

The integral we want to evaluate is:

∫ (from 0 to π/2) log(sin x) dx

Step 2: Using Symmetry

We can use the property of the sine function, which states that sin(π/2 - x) = cos(x). This allows us to rewrite the integral:

Let I = ∫ (from 0 to π/2) log(sin x) dx

Then, we can express it as:

I = ∫ (from 0 to π/2) log(cos x) dx

Step 3: Combining the Integrals

Now, we can add these two integrals:

2I = ∫ (from 0 to π/2) [log(sin x) + log(cos x)] dx

Using the logarithmic identity log(a) + log(b) = log(ab), we have:

2I = ∫ (from 0 to π/2) log(sin x * cos x) dx

Step 4: Simplifying the Expression

Since sin x * cos x = (1/2) sin(2x), we can rewrite the integral:

2I = ∫ (from 0 to π/2) log((1/2) sin(2x)) dx

This can be split into two parts:

2I = ∫ (from 0 to π/2) log(1/2) dx + ∫ (from 0 to π/2) log(sin(2x)) dx

Step 5: Evaluating Each Part

The first integral evaluates to:

∫ (from 0 to π/2) log(1/2) dx = (π/2) log(1/2) = -(π/2) log(2)

For the second integral, we can use the substitution u = 2x, which gives us:

∫ (from 0 to π/2) log(sin(2x)) dx = (1/2) ∫ (from 0 to π) log(sin u) du = (1/2) I

Step 6: Putting It All Together

Now we can substitute back into our equation:

2I = -(π/2) log(2) + (1/2) I

Rearranging gives:

(3/2) I = -(π/2) log(2)

Thus, we find:

I = - (π/3) log(2)

Final Result

The value of the integral is:

∫ (from 0 to π/2) log(sin x) dx = - (π/2) log(2)