InterviewSolution
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InterviewSolution
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`int_(0)^((pi)/(2))(2log sin x- log sin 2x)dx` |
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Answer» माना `" "I=int_(0)^(pi//2)(2log sin x-log sin 2x)dx` `=int_(0)^(pi//2)(log sin^(2)x-log sin 2x)dx` `=int_(0)^(pi//2)log((sin^(2)x)/(sin2x))dx` `=int_(0)^(pi//2)log((sin^(2)x)/(2sin x cos x))dx` `=int_(0)^(pi//2)log((tanx)/(2))dx` `=int_(0)^(pi//2)log(tanx)-log 2dx` `=int_(0)^(pi//2)log(tanx)dx-int_(0)^(pi//2)log2dx` `rArr" "I=I_(1)-log2[x]_(0)^((pi)/(2))=I_(1)-((pi)/(2)-0)log 2" ...(1)"` जहाँ,`" "I_(1)=int_(0)^(pi//2)log(tan x)dx" ...(2)"` `rArr" "I_(1)=int_(0)^(pi//2)log[tan((pi)/(2)-x)]dx` `[because int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx]` `rArr" "I_(1)=int_(0)^(pi//2)log(cotx)dx" ...(3)"` समीकरण (2 ) और (3 ) को जोड़ने पर , `2I_(1)=int_(0)^(pi//2){(log(tanx)+log(cotx)}dx` `int_(0)^(pi//2)log(tanxcotx)dx` `=int_(0)^(pi//2)log 1dx=0` `rArr" "I_(1)=0` `I_(1)` का मान समीकरण (1 ) में रखने पर, `I=0-(pi)/(2)log 2=-(pi)/(2)log 2` |
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