Exams › JEE Advanced › Maths
Correct answer: pi/2 - 1
Using the substitution property with f(a+b-x) = f(-x): Let I1 = integral from -pi/2 to pi/2 of x*sin(x)/(1+e^x) dx and I2 = integral from -pi/2 to pi/2 of (-x)*sin(-x)/(1+e^(-x)) dx. Note sin(-x) = -sin(x), so I2 = integral of x*sin(x)/(1+e^(-x)) dx = integral of x*sin(x)*e^x/(1+e^x) dx. Therefore 2I = I1 + I2 = integral of x*sin(x)*[1/(1+e^x) + e^x/(1+e^x)] dx = integral of x*sin(x) dx from -pi/2 to pi/2. Since x*sin(x) is even: = 2 * integral from 0 to pi/2 of x*sin(x) dx. Integrate by parts: [-x*cos(x) + sin(x)] from 0 to pi/2 = [0 + 1] - [0 + 0] = 1. So 2I = 2*1 = 2, giving I = 1. Wait: recalculating: 2*integral from 0 to pi/2 of x*sin(x) dx = 2*[-x cos(x) + sin(x)] from 0 to pi/2 = 2*[(-pi/2 * 0 + 1) - (0 + 0)] = 2*1 = 2. So 2I = 2, I = 1. But checking option: the answer 1 is option (3). Actually let me recheck the options mapping. Given options are pi/2 - 1, pi - 2, 1, pi/4. The correct answer from the calculation is I = 1.