The defined integral below is, in fact, very hard to evaluate by common means.
I = ∫_{o}^{π/2} √sin(x)/(√sin(x)+√cos(x)) dx
However, if you make the substitution x=(π/2y), it becomes surprisingly easy to solve, by applying a basic concept of "defined integrals".
With this hint, can you, now, evaluate its value?
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 20050626 06:14:43 