Integrate ln(sin(x)) from 0 to π/2.

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

Let's start by setting u = ln(sin(x)) and dv = dx. Then, we can differentiate u and integrate dv as follows:

du = (1/sin(x)) * cos(x) dx,
v = x.

Using the integration by parts formula:

∫u dv = uv - ∫v du,

we have:

∫ln(sin(x)) dx = x * ln(sin(x)) - ∫x * (1/sin(x)) * cos(x) dx.

To evaluate the integral on the right-hand side, we need to simplify the integrand further.

Notice that x * (1/sin(x)) * cos(x) can be simplified as x * cot(x) * cos(x).

Using the identity cot(x) = cos(x)/sin(x), we can rewrite the integrand as x * cos(x) * (cos(x)/sin(x)) = x * cos^2(x) / sin(x).

Now, let's evaluate the integral of x * cos^2(x) / sin(x):

∫x * cos^2(x) / sin(x) dx.

We can use a substitution to simplify this integral. Let u = sin(x), then du = cos(x) dx.

The integral becomes:

∫x * cos^2(x) / sin(x) dx = ∫x * (1 - sin^2(x)) du.

Expanding the integral:

∫x * (1 - sin^2(x)) du = ∫(x - x * sin^2(x)) du.

Now, we can integrate each term separately:

∫x du = x * u = x * sin(x),

and

∫-x * sin^2(x) du = -x * ∫sin^2(x) du.

The integral of sin^2(x) can be evaluated using the double-angle formula for cosine: cos(2x) = 1 - 2sin^2(x).

sin^2(x) = (1 - cos(2x))/2.

Substituting this back into the integral:

∫-x * sin^2(x) du = -x * ∫(1 - cos(2x))/2 du
= -x/2 * [u - (1/2) * sin(2x)].

Now, let's substitute u = sin(x) back into the expression:

= -x/2 * [sin(x) - (1/2) * sin(2x)].

Therefore, the integral of ln(sin(x)) from 0 to π/2 is given by:

∫ln(sin(x)) dx = x * ln(sin(x)) - x/2 * [sin(x) - (1/2) * sin(2x)].

Now, we can evaluate the definite integral by substituting the limits of integration:

∫[0 to π/2] ln(sin(x)) dx
= [(π/2) * ln(sin(π/2)) - (π/2)/2 * [sin(π/2) - (1/2) * sin(π)]]

[(0 * ln(sin(0)) - 0/2 * [sin(0) - (1/2) * sin(0)])].
Simplifying further:

= (π/2) * ln(1) - (π/4) * [1 - 0]

(0 * ln(0) - 0/2 * [0 - (1/2) * 0]).
Since ln(1) = 0 and sin(0) = 0, the