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/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?

 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