InterviewSolution
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InterviewSolution
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`I=(2)/(pi) int_(-pi//4)^(pi//4) (dx)/((1+e^(sinx))(2-cos2x))` then find `27I^(2)` |
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Answer» Correct Answer - 4 `I=(2)/(pi)underset(-(pi)/(4))overset(pi//4)int(dx)/((1+e^(sinx))(2-cos2x))` ..(i) by a+b-x property `I=(2)/(pi)underset(-(pi)/(4))overset(pi//4)int(dx)/((1+e^(-sinx))(2-cos2x))=(2)/(pi)overset(-(pi)/(4))overset(pi//4)int(e^(sinx)dx)/((1+e^(sinx))(2-cos2x))dx` ...(ii) adding (1) and (2) `2I=(2)/(pi)underset(-(pi)/(4))overset(pi//4)int((1+e^(sinx)))/((1+e^(sinx))(2-cos2x))dximpliesI=(1)/(pi)underset(-(pi)/(4))overset(pi//4)int(1)/(2(2cos^(2)x-1))dx=(1)/(pi)underset(-(pi)/(4))overset(pi//4)(sec^(2)x)/(3sec^(2)x-2)dx` put `tanx=t.sec^(2)xdx=dt` `=(2)/(pi)int_(0)^(1)(dt)/(3t^(2)+1)=(2)/(3pi)(1)/(((1)/(sqrt(3))))(tan^(-1)((t)/(1//sqrt(3))))_(0)^(1)=(2)/(3pi)(tan^(-1)(sqrt(3))-tan^(-1)(0))=(2)/(sqrt(3)pi)((pi)/(3))=(2)/(3sqrt(3))` Now `27I^(2)=27xx(4)/(27)=4` |
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