ज्ञात करे [ Evaluate ] ` int _(0)^(pi//2) (sin^(2) x ) /(sin x + cos x ) dx `

Explanation

माना कि `I = int _(0)^(pi//2) (sin^(2)x)/(sin x+ cos x ) dx `
तो ` I = int _(0)^(pi//2) (sin^(2)(pi/2-x))/(sin (pi/2-x) + cos x (pi/2 - x))dx `
`[ :' int _(0)^(2) f (x) dx = int _(0)^(a) f(a-x) dx ] `
या ` I = int _(0)^(pi//2) (cos^(2) x)/(cos x + sin x ) dx `
(1 ) और ( 2 ) को जोड़ने पर हमे मिलता है ,
` 2I =int _(0)^(pi//2) [ (sin^(2) x)/(sin x + cos x )+ (cos^(2)x)/(sin x + cos x)] dx =int _(0)^(pi//2) 1/(sinx + cos x)dx `
` rArr 2I = int _(0)^(pi//2) 1/((2tan x//2)/(1+tan^(2) x//2 )+ (1-tan^2x//2)/(1+ tan^(2) x//2))dx`
`rArr 2I = int _(0)^(pi//2) (1+tan^(2). x/2)/ (2 tan. x/2 + 1- tan^(2) . x/2 )dx = int_(0)^(pi//2) (sec^(2). x/2 ) /(2 tan. x/2 + 1- tan^(2). x/2 ) dx ` ...(3)
`tan. x/2 = z " रखे, तो " sec^(2) . x/2 .1/2 dx = dz rArr sec^(2) . x/2 dx = 2 dz `
साथ ही जब `x = 0 , z = tan 0 = 0 ` तथा जब ` x = pi/2 , z = tan. pi/4 =1 `
अब ( 3 ) से ` 2I = int _(0)^(1) (2dz)/(2z + 1- z^(2)) = 2 int _(0)^(1) 1/((sqrt2)^(2) - (z- 1)^(2)) dz`
` = 2 xx 1/(2sqrt(2)) [ log | (sqrt(2) +z -1 )/(sqrt(2) - z +1 )|]_(0)^(1)`
` = 1/(sqrt(2)) { log ((sqrt(2))/(sqrt(2))) - log ((sqrt(2)-1)/(sqrt(2)+1))} = 1/sqrt(2) { 0 -log ((sqrt(2)-1)/(sqrt(2)+1))}`
`= - 1/(sqrt(2)) log { (sqrt(2)-1)^(2)/((sqrt(2)+1) (sqrt(2)-1))}= - 1/(sqrt(2)) log . (sqrt(2)-1)" " - 2/(sqrt(2)) log (sqrt(2)-1)`
` :. I = -1/(sqrt(2)) log (sqrt(2)-1) = 1/(sqrt(2)) log (sqrt(2) +1)" " [ :' sqrt(2) -1 = 1/(sqrt(2)+1)]`