Integrate log (sin x) from 0 to π/2.

To integrate the function log(sin x) from 0 to π/2, we can use integration by parts.

Let's denote the integral as I:
I = ∫[0,π/2] log(sin x) dx

Integration by parts states that ∫u dv = uv - ∫v du, where u and v are functions.

To apply integration by parts, we need to choose u and dv. In this case, we'll let:
u = log(sin x) => du = (1/sin x) * cos x dx
dv = dx => v = x

Now we can substitute these values into the integration by parts formula:

I = ∫[0,π/2] u dv
= [uv] - ∫[0,π/2] v du
= [x * log(sin x)] - ∫[0,π/2] x * (1/sin x) * cos x dx

Next, we evaluate each term separately:

[x * log(sin x)] evaluated from 0 to π/2:
= (π/2 * log(sin π/2)) - (0 * log(sin 0))
= (π/2 * log(1)) - (0 * log(0))
= (π/2 * 0) - (0 * -∞)
= 0

Now let's evaluate the integral term:

∫[0,π/2] x * (1/sin x) * cos x dx

We can simplify this expression by canceling out sin x and cos x:

∫[0,π/2] x * (1/sin x) * cos x dx
= ∫[0,π/2] x dx
= [x^2/2] evaluated from 0 to π/2
= (π^2/8) - (0^2/2)
= π^2/8

Putting it all together:

I = [x * log(sin x)] - ∫[0,π/2] x * (1/sin x) * cos x dx
= 0 - (π^2/8)
= -π^2/8

Therefore, the definite integral of log(sin x) from 0 to π/2 is -π^2/8.