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