InterviewSolution
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InterviewSolution
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Evaluate `int_(0)^(pi//2)(x)/((sinx+cosx))dx`. |
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Answer» Let `I=int_(0)^(pi//2)(x)/((sinx+cosx))dx`……..`(i)` Then, `I=int_(0)^(pi//2)(((pi)/(2)-x))/(sin[(pi//2)-x]+cos[(pi//2)-x])dx` or `I=(((pi)/(2)-x))/((cosx+sinx))dx=int_(0)^(pi//2)(((pi)/(2)-x))/((sinx+cosx))dx`.......`(ii)` Adding `(i)` and `(ii)`, we get `2I=(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-tam^(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` `=(pi)/(4)int_(0)^(1)(2dt)/((1-t^(2)+2t))`, where `t=tan.(x)/(2)` [`x=0impliest=0` and `x=(pi)/(2)impliest=1`] `=(pi)/(2)int_(0)^(1)(dt)/([(sqrt(2))^(2)-(t-1)^(2)])dt` `=(pi)/(2)*(1)/(2sqrt(2))log|(sqrt(2)+(t-1))/(sqrt(2)-(t-1))|_(0)^(1)=(pi)/(4sqrt(2))log|(sqrt(2)+1)/(sqrt(2)-1)|`. |
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