ज्ञात करे ` int _(0)^(pi)x/(a^(2) cos^(2) x + b^(2) sin ^(2) x) dx`

Explanation

माना कि ` I = int _(0)^(pi) x/(a^(2) cos ^(2) x + b^(2) sin^(2) x ) dx " "`…(1)
तो ` I = int _(0)^(pi) ((pi-x)) / (a^(2) cos (pi-x) = b^(2) sin^(2) (pi-x) ) dx `
` [ :' int _(0)^(a) f(x) dx= int _(0)^(a) f(a- x) dx ] `
या , ` I = int _(0)^(pi) (pi-x)/(a^(2) cos^(2) x+b^(2) sin^(2) x )dx" "` ...(2)
(1 ) और (2) को जोड़ने पर हमे मिलता है ,
` 2I = int _(0)^(pi) (x+pi-x)/(a^(2) cos^(2) x+b^(2) sin^(2) x )dx `
` = pi int _(0)^(pi) 1/(a^(2) cos x +b^(2) sin^(2) x )dx`
` = 2pi int _(0)^(pi//2) 1/(a^(2) cos x +b^(2) sin^(2) x )dx`
` [ :' int _(0)^(2a) f(x) dx = 2 int _(0)^(a) f(x) dx , " यदि " f(2a - x) = f(x)]`
` :. 2I = 2pi int _(0)^(pi//2) (sec^(2)x)/(a^(2) +b^(2) tan^(2)x)` [ अंश तथा हर को `cos^(2)x` से भाग देने पर ]
या , ` I = pi int _(0)^(pi//2) (sec^(2)x)/(a^(2) +b^(2) tan^(2)x)dx`
माना कि `tan x = z ` तो , ` sec^(2)"x dx" = dz`
साथ ही ` x = 0 rArr z = tan 0 = 0 " तथा " x to pi/2 " या " z to oo`
` :. I = pi int _(0)^(oo) (dz)/(a^(2)+b^(2)z^(2))= pi/(b^(2)) int _(0)^(oo) (dz)/((a//b)^(2) +z^(2)) = pi/(b^(2)) xx 1/(a//b) [ tan^(-1) (x/(a//b))]_(0)^(oo)`
`rArr I = pi/(ab) [ tan^(-)((bz)/a)]_(0)^(oo) = (tan^(-1) oo - tan^(-1) 0) = pi/(ab) (pi/ 2 - 0) = (pi^(2))/(2ab)`