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For x in (0, pi/2), evaluate I = integral from 0 to sin²(x) of arcsin(sqrt(t)) dt + integral from 0 to cos²(x) of arccos(sqrt(t)) dt.

1. pi/2
2. 1
3. pi/4
4. None of these

Correct answer: pi/4

Solution

Let I(x) = I1 + I2. By Leibniz rule: dI1/dx = arcsin(sqrt(sin²(x))) * d(sin²x)/dx = x * sin(2x). dI2/dx = arccos(sqrt(cos²(x))) * d(cos²x)/dx = x * (-sin(2x)). So dI/dx = 0 for all x in (0,pi/2). I is constant. Evaluate at x=pi/4: I = integral from 0 to 1/2 of [arcsin(sqrt(t))+arccos(sqrt(t))] dt = integral from 0 to 1/2 of (pi/2) dt = pi/4.

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