Integrate [sinx/(1+cos^2(x))] from pi/2 to pi.

To integrate the given expression, we can use the substitution method. Let's use the substitution u = cos(x), so we have du = -sin(x) dx.

Now, let's rewrite the expression in terms of u:
sin(x) = √(1 - cos^2(x)) = √(1 - u^2)
dx = -du/sin(x) = -du/√(1 - u^2)

Substituting these into the integral, it becomes:
∫ [sin(x) / (1 + cos^2(x))] dx = ∫ [(√(1 - u^2) / (1 + (1 - u^2))^2] * (-du/√(1 - u^2))

Simplifying this expression, we get:
∫ [(√(1 - u^2) / (2 - u^2)] * (-du/√(1 - u^2)) = -∫ du / (2 - u^2)

Now we can integrate the expression:
-∫ du / (2 - u^2) = - (1/2) ∫ du / (1 - (u^2/2))

To integrate this expression, we can use the partial fraction decomposition method.

The expression 1 / (1 - (u^2/2)) can be written as A / (1 - u√2) + B / (1 + u√2), where A and B are constants to be determined.

Multiplying both sides by (1 - u√2) * (1 + u√2):
1 = A * (1 + u√2) + B * (1 - u√2)

Now, let's solve for A and B. Setting u√2 = 1:
1 = A * (1 + 1) + B * (1 - 1)
1 = 2A

Setting u√2 = -1:
1 = A * (1 - 1) + B * (1 + 1)
1 = 2B

So we have A = 1/2 and B = 1/2.

Now, let's rewrite the integral:
-(1/2) ∫ du / (1 - (u^2/2)) = -(1/2) ∫ [(1/2) / (1 - u√2) + (1/2) / (1 + u√2)] du

Integrating these individual terms, we get:
-(1/2) * (1/2) * ln|1 - u√2| - (1/2) * (1/2) * ln|1 + u√2| + C

Substituting back u = cos(x):
= -(1/4) * ln|1 - cos(x)√2| - (1/4) * ln|1 + cos(x)√2| + C

Now, let's evaluate the definite integral from π/2 to π:
-(1/4) * ln|1 - cos(π)√2| - (1/4) * ln|1 + cos(π)√2| - (-(1/4) * ln|1 - cos(π/2)√2| - (1/4) * ln|1 + cos(π/2)√2|

Now we can simplify this expression:

-(1/4) * ln|1 - (-1)√2| - (1/4) * ln|1 + (-1)√2| - (-(1/4) * ln|1 - 0√2| - (1/4) * ln|1 + 0√2|

= -(1/4) * ln|1 + √2| - (1/4) * ln|1 - √2| - 0 - 0

= -(1/4) * ln|1 + √2| + (1/4) * ln|1 - √2|

Therefore, the value of the definite integral of sin(x) / (1 + cos^2(x)) from π/2 to π is:
-(1/4) * ln|1 + √2| + (1/4) * ln|1 - √2|