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

Explanation

माना कि ` f(x) = x/(sin x + cos x) " " ` … (1)
तो , ` f(pi/2-x) = (pi/2-x)/(sin(pi/2-x)+cos(pi/2-x))`
या , ` f (pi/2 - x) = (pi/2-x)/(cos x +sinx) " "` (2)
अब ,(1) +(2) `rArr f(x) +f(pi/2-x) = pi/2 1/(2cos x+ sinx)`
` = pi/(2sqrt(2)cos (x-pi/4))=pi/(2sqrt(2)) sec (x- pi/4)`
अब , ` I = 1/2 int_(0)^(pi//2) [ f(x)+f(pi/2-x)]dx `
` =1/2 . pi/(2sqrt(2)) int_(0)^(pi//2) sec (x- pi/4) dx `
` = pi/(4sqrt(2))[log|sec(x-pi/4)+tan (x-pi/4)|]_(0)^(pi//2)`
` = pi/(4sqrt(2)) [ log | sec. pi/4 +tan pi/4 | - log |sec.pi/4 - tan .pi/4 |]`
` = pi/(4sqrt(2)) [ log (sqrt(2)+1)-log (sqrt(2)-1)] `
` = pi/(4sqrt(2)) log ((sqrt(2)+1)/(sqrt(2)-1)) = pi/(4sqrt(2))log (sqrt(2)+1)^(2) = pi/(2sqrt(2)) log (sqrt(2)+1)`