ज्ञात करे ।[ Evaluate ] `int_(0)^(pi) (x sin x)/(1+cos ^(2)x)dx`

Explanation

माना कि ` I = int _(0)^(pi) (x sin x)/(1+cos^(2)x)dx" "` …(1)
तो ` I = int _(0)^(pi) ((pi-x)sin (pi-x))/(1+cos^(2)(pi-x))" " [ :' int _(0)^(a) f(x) dx = int _(0)^(a) f(a-x) dx]`
या , `I = int_(0)^(pi) ((pi-x)sinx)/(1+cos^(2)x) dx" "` ...(2)
(1) और (2) को जोड़ने पर हमे मिलता है ,
` 2I = int _(0)^(pi) ((x+pi-x))/(1+cos^(2)x)dx = pi int _(0)^(pi) (sin x)/(1+cos^(2)x) dx `
` rArr I = pi/2 int _(0)^(pi) (sinx)/(1+cos^(2)x)dx `
` z = cos x " रखे ,तो " dx = - sin x dx `
साथ ही जब ` x = 0 , z = cos 0 = 1`
तथा जब ` x = pi, z = cos pi =-1`
(3) से , `I = pi/2 int_(1)^(-1) (-dz)/(1+z^(2)) = -pi/2 int_(1)^(-1) 1/(1+z^(2)) dz = -pi/2 [ tan^(-1)z]_(1)^(-1)`
`= pi/2 [ tan^(-1) (-1) - tan^(-1) (1)] = -pi/2 [ -pi/4 -pi/4 ] = (pi^(2))/4`
अतः ` I = (pi^(2))/4`