InterviewSolution
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InterviewSolution
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`int_(0)^(pi//2)(sqrt(cosx))/((sqrt(cosx)+sqrt(sinx)))dx=?` |
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Answer» Let `I=int_(0)^(pi//2)(sqrt(cosx))/((sqrt(sinx)+sqrt(cosx)))dx`………..`(i)` Using the result `int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx` in `(i)`, we get `I=int_(0)^(pi//2)(sqrt(cos[(pi//2)-x]))/(sqrt(sin[(pi//2)-x])+sqrt(cos[(pi//2)-x]))dx` or `I=int_(0)^(pi//2)(sqrt(sinx))/(sqrt(cosx)+sqrt(sinx))dx=int_(0)^(pi//2)(sqrt(sinx))/(sqrt(sinx)+sqrt(cosx))dx`.........`(ii)` Adding `(i)` and `(ii)` we get `2I=int_(0)^(pi//2)(sqrt(cosx))/((sqrt(sinx)+sqrt(cosx)))dx+int_(0)^(pi//2)(sqrt(sinx))/((sqrt(sinx)+sqrt(cosx)))dx` `=int_(0)^(pi//2)((sqrt(sinx)+sqrt(cosx)))/((sqrt(sinx)+sqrt(cosx)))dx=int_(0)^(pi//2)dx=[x]_(0)^(pi//2)=(pi)/(2)`. `:.I=(pi)/(4)`, i.e., `int_(0)^(pi//2)(sqrt(cosx))/((sqrt(sinx)+sqrt(cosx)))dx=(pi)/(4)`. |
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