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An ingenious evaluation (Posted on 2005-06-26) Difficulty: 3 of 5
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)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution 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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