Find the value of the Integral,integral of limits -pi/2 to pi/2 (root ofcos x - cos3 x)dx

solution: I = Integral (-pi/2 to pi/2) [cos x - cos 3x]^1/2 .dx = integral (-pi/2 to pi/2) [cos x - (4cos3 x - 3cos x)]^1/2 .dx ; cos 3x = 4cos3 x - 3cos x = integral (-pi/2 to pi/2) [ 4cox x - 4 cos^3 x ]1/2 .dx = integral (-pi/2 to pi/2) [ 4cox x {1 - cos^2 x} ]1/2 .dx = integral (-pi/2 to pi/2) [ 4cox x sin^2 x ]1/2 .dx = integral (-pi/2 to pi/2) [ 2 sin x {cox x }^1/2 ] .dx substituting cos x = t so ; sin x .dx = - dt lower limit: sin (-pi/2) = -1 upper limit: sin (pi/2) = 1 I = integral (-1 to 1) [ 2{ t}^1/2 ] .dt = 4/3 [1+ i] : here i = iota Thanks and Regards, Ajay verma, askIITians faculty, IIT HYDERABAD