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Evaluate the definite integral from pi/2 to 5*pi/2 of [e^(arctan(sin x)) / (e^(arctan(sin x)) + e^(arctan(cos x)))] dx. If the value equals k*pi where k is a natural number, find k.

1. 1
2. 2
3. 3
4. 4

Correct answer: 1

Solution

Let I = integral from pi/2 to 5*pi/2 of f(x) dx. The integrand has period 2*pi. Note sin(x+pi) = -sin(x) and cos(x+pi) = -cos(x). Then f(x+pi) = e^(arctan(-sin x)) / (e^(arctan(-sin x)) + e^(arctan(-cos x))) = e^(-arctan(sin x)) / (e^(-arctan(sin x)) + e^(-arctan(cos x))). Multiplying numerator and denominator by e^(arctan(sin x)+arctan(cos x)): f(x+pi) = e^(arctan(cos x)) / (e^(arctan(sin x)) + e^(arctan(cos x))). So f(x) + f(x+pi) = 1. Therefore I = integral_(pi/2)^(5pi/2) f(x) dx = integral_(pi/2)^(3pi/2) [f(x) + f(x+pi)] dx = integral_(pi/2)^(3pi/2) 1 dx = pi. So k = 1.

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