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Correct answer: pi² / 4
Let I = integral[0 to pi] x*tan(x)/(sec(x)+cos(x)) dx. Using the property integral[0 to a] f(x)dx = integral[0 to a] f(a-x)dx, substitute x -> pi-x. tan(pi-x) = -tan(x), sec(pi-x)+cos(pi-x) = -sec(x)-cos(x). So f(pi-x) = (pi-x)*(-tan x)/(-sec x - cos x) = (pi-x)*tan(x)/(sec x + cos x). Adding I + I = pi * integral[0 to pi] tan(x)/(sec x+cos x)dx. The integrand tan(x)/(sec x + cos x) = sin(x)*cos(x)/(1+cos²(x)). Let u = cos x, du = -sin x dx, limits pi->-1, 0->1. Integral becomes integral[-1 to 1] u/(1+u²) * (-du/(-1))... Actually using symmetry the integral from 0 to pi of sin(x)cos(x)/(1+cos²(x)) dx = 2 * integral from 0 to pi/2 of sin(x)cos(x)/(1+cos²(x)) dx. Let u = cos x: integral from 1 to 0 of u/(1+u²)*(-du) = integral[0 to 1] u/(1+u²) du = (1/2)*ln(2). So 2I = pi*(1/2)*ln(2)? That gives a log answer. Let me redo: tan(x)/(sec(x)+cos(x)) = [sin(x)/cos(x)] / [1/cos(x)+cos(x)] = sin(x)/(1+cos²(x)). So 2I = pi * integral[0 to pi] sin(x)/(1+cos²(x)) dx. Let u=cos x: = pi*integral[-1 to 1] du/(1+u²) = pi*[arctan(u)]₋₁¹ = pi*(pi/4 - (-pi/4)) = pi*(pi/2) = pi²/2. Therefore I = pi²/4.