माना कि ` I = int _(-pi)^(pi) (2x (1+ sin x))/(1+cos^(2) x)dx `

Explanation

तो , ` I = int _(pi)^(pi) (2x)/(1+cos^(2) x ) dx + int _(-pi)^(pi) (2x sin x)/(1+cos^(2) x ) dx = I _(1) + I_(2)`
जहाँ ` I _(1) = int _(-pi)^(pi) (2x)/(1+cos^(2) x ) dx " तथा " I_(2) = int _(-pi)^(pi) (2x sin x )/(1+cos^(2) x) dx " "` …(2)
चूँकि `(2x)/(1+cos^(2)x) ` एक विषम फलन है तथा ` (2x sin x)/(1+cos^(2)x ) dx ` एक सम फलन है ।
` :. I _(1) = 0 " " I_(2) = 2 int _(0)^(pi) (2x sin x)/(1+cos^(2) x ) dx `
अब , ` I _(2) = 4 int _(0)^(pi) (x sin x)/(1+cos^(2) x ) dx " "` ..(3) ,
` :. I _(2) = 4 int _(0)^(pi) ((pi-x)sin (pi-x))/(1+cos^(2) (pi-x))dx`
(3 ) और (4 ) को जोड़ने पर हमे मिलता है ,
` 2I_(2) = 4pi int _(0)^(pi) 1/(1+cos^(2) x ) sin xdx = - 4x int_(1)^(-1) 1/(1+t^(2)) dt , " जहाँ " t = cos x `
` = - 4pi [ tan^(-1) t ] _(1)^(-1) = - 4pi [ -pi//4 - pi//4 ] = 2pi^(2)`
` :. 2I_(2) = 2pi^(2) rArr I_(2) = pi^(2) `
अतः ( 2 ) से , ` I = 0 + pi^(2) = pi^(2) `