perplexus.info :: Calculus : An ingenious evaluation

An ingenious evaluation (Posted on 2005-06-26)

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?

Submitted by pcbouhid

Rating: 2.0000 (4 votes)

Solution: (Hide)

I = (Int from 0º to pi/2) of [sqrt(sin x) / (sqrt(sin x) + sqrt(cos x))]dx......(1)

Let's make the substitution suggested, x = (pi/2 - y).

Thus, the new limits are : for x=0º, y=pi/2, and for x=pi/2, y=0º. Also :

sin x = sin(pi/2 - y) = cos y

cos x = cos(pi/2 - y) = sin y

And, dx/dy = -1 ==> dx = -dy.

The integral becomes :

I = (Int from 0º to pi/2) of [sqrt(cos y) / (sqrt(cos y) + sqrt(sin y))]dy.......(2)

Since one DEFINITE integral IS NOT function of the variable, but a funtion of their limits, we can change the variable of integral (2) to...x :

I = (Int from 0º to pi/2) of [sqrt(cos x) / (sqrt(cos x) + sqrt(sin x))]dx.......(3)

Now, summing up (1) and (3):

2*I = (Int fom 0º to pi/2)dx

2*I = x(from 0º to pi/2) = pi/2 - 0º = pi/2.

Finally:

I = pi/4

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
TO: LEVIK (2)	K Sengupta	2005-11-12 02:44:55
TO : LEVIK (1)	K Sengupta	2005-11-12 02:39:39
Full Solution	K Sengupta	2005-11-12 02:31:36
Problem	K Sengupta	2005-07-01 07:22:58
Problem	K Sengupta	2005-07-01 07:22:56
Easy	Rex	2005-06-26 14:02:46
Solution	Bractals	2005-06-26 06:14:43

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