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Evaluate : `int_(0)^(pi) (x sin x )/(1 + sin x ) dx ` |
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Answer» Let `underset(0)overset(pi) int (x sin x dx)/(1 + sin x ) = I` ....(i) then, ` I =underset(0)overset(pi) int((pi -x)sin (pi -x))/(1+sin (pi -x))dx ` ` = underset(0)overset(pi) int ((pi -x)sin x)/(1 + sin x) dx ` ....(ii) Adding equations (i) and (ii), `2 I = underset(0)overset(pi) int (pi sin x dx)/(1 + sin x ) ` or ` I = pi/2 underset(0)overset(pi) int (sin x)/(1 + sin x) dx ` or ` I = pi/2 underset(0)overset(pi) int((1-sin x) sin x)/((1+sin x)(1-sin x)) dx` or ` I = pi/2 underset(0)overset(pi) int (sin x - sin^(2) x)/(1- sin^(2) x) dx ` or ` I = pi/2 underset(0)overset(pi) int (sin x - sin^(2)x)/(cos^(2)x) dx ` or `I = pi/2 underset(0)overset(pi)int(sin x )/(cos^(2) x) - pi/2 underset(0)overset(pi) int (sin^(2)x)/(cos ^(2) x ) dx ` or ` I = pi/2 underset(0)overset(pi) int tan x sec x dx - pi/2 underset(0)overset(pi)int (1-cos^(2)x)/(cos^(2) x) dx ` or `I = pi/2 [sec x]_(0)^(pi) - pi/2 underset(0)overset(pi)intsec^(2) x dx + pi/2 underset(0)overset(pi)int dx ` or ` I = pi/2 [sec pi - sec 0] - pi/2 [tan x]_(0)^(pi) + pi/2 [x]_(0)^(pi) ` or ` I = pi/2 [-1-1] - pi/2 [tan pi - tan0] + pi/2 [pi - 0]` or ` I = - pi - pi/2 [0-0] + pi/2[pi]` or ` I = pi [pi/2 -1]` Thus, `underset(0)overset(pi)int (x sin x dx)/(1 + sin x ) = pi [(pi -2)/2]` |
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