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 An ingenious evaluation (Posted on 2005-06-26)
The defined integral below is, in fact, very hard to evaluate by common means.

I = ∫oπ/2 √sin(x)/(√sin(x)+√cos(x)) dx

However, if you make the substitution x=(π/2-y), it becomes surprisingly easy to solve, by applying a basic concept of "defined integrals".

With this hint, can you, now, evaluate its value?

 See The Solution Submitted by pcbouhid Rating: 2.0000 (4 votes)

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 Solution | Comment 1 of 7
` `
` I = integral(sqrt(sin(x))/(sqrt(sin(x))+sqrt(cos(x))) dx; 0 to PI/2)`
`Substitution x=(PI/2 - y),`
` I = integral(sqrt(sin(PI/2 - y))/(sqrt(sin(PI/2 - y))+             sqrt(cos(PI/2 - y))) d(PI/2 - y); PI/2 to 0)`
`   = integral(sqrt(cos(y))/(sqrt(cos(y))+sqrt(sin(y))) -dy; PI/2 to 0)`
`   = integral(sqrt(cos(x))/(sqrt(cos(x))+sqrt(sin(x))) dx; 0 to PI/2)`
`Add the original,`
`2I = integral(sqrt(cos(x))/(sqrt(cos(x))+sqrt(sin(x))) dx; 0 to PI/2) +`
`     integral(sqrt(sin(x))/(sqrt(sin(x))+sqrt(cos(x))) dx; 0 to PI/2)`
`   = integral(sqrt(cos(x))/(sqrt(cos(x))+sqrt(sin(x))) +`
`              sqrt(sin(x))/(sqrt(sin(x))+sqrt(cos(x))) dx; 0 to PI/2)`
`   = integral((sqrt(cos(x))+sqrt(sin(x)))/(sqrt(cos(x))+sqrt(sin(x))) dx;`
`              0 to PI/2)`
`   = integral( dx; 0 to PI/2)`
`   = PI/2`
`Therefore,`
` I = PI/4`
` `

 Posted by Bractals on 2005-06-26 06:14:43

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