InterviewSolution
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InterviewSolution
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`int_(0)^(pi//2)(x)/((sinx+cosx))dx` का मान ज्ञात कीजिए । |
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Answer» माना `I=int_(0)^(pi//2)(x)/((sinx+cosx))dx" …(1)"` ` I=int_(0)^(pi//2)(((pi)/(2)-x))/(sin[(pi//2)-x]+cos[(pi//2)-x])dx` `I=int_(0)^(pi//2)(((pi)/(2)-x))/((cosx+sinx))dx` `=int_(0)^(pi//2)(((pi)/(2)-x))/((sinx+cosx))dx" ...(2)"` अब, समीकरण (1 ) व (2 ), को जोड़ने पर `@I=(pi)/(2)int_(0)^(pi//2)(dx)/((sinx+cosx))` `I=(pi)/(4)int_(0)^(pi//2)(dx)/((sinx+cosx))` `=(pi)/(4)int_(0)^(pi//2)(dx)/([(2tan(x//2))/(1+tan^(2)(x//2))+(1-tan^(2)(x//2))/(1+tan^(2)(x//2))])` `=(pi)/(4)int_(0)^(pi//2)(sec^(2)(x//2))/(1-tan^(2)(x//2)+2tan(x//2))dx` यदि `tan.(x)/(2)=t rArr sec^(2).(x)/(2)dx=2dt` तथा सीमाएँ t = 0 पर वx = 0 पर `x=pi//2,` तब `=(pi)/(4)int_(0)^(i//2)(sec^(2)(x//2))/(1-tan^(2)(x//2)+2tan(x//2))dx` `=(pi)/(4)int_(0)^(1)(2dt)/(1-t^(2)+2t)` `=(pi)/(2)int_(0)^(1)(dt)/([(sqrt2)^(2)-(t-1)^(2)])` `=(pi)/(2).(1)/(2sqrt2)log[(sqrt2+(t-1))/(sqrt2-(t-1))]_(0)^(1)` `=(pi)/(4sqrt2)log[(sqrt2+1)/(sqrt2-1)]` |
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