Re: Analysis with sinx/sinx+cosx.

On Sun, 20 Apr 2008 17:46:09 +0900, "mina_world"
<mina_world@xxxxxxxxxxx> wrote:

Hello teacher~

int{0 to pi/2} (sin x) / [(sin x) + (cos x)] dx = ?

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Let x = pi/2 - t.

sin x = cos t
cos x = sin t
dx = - dt

so, int{0 to pi/2} (sin x) / [(sin x) + (cos x)] dx
= int{pi/2 to 0} (cos t) / [(cos t) + (sin t)] (-dt)
= int{0 to pi/2} (cos t) / [(cos t) + (sin t)] dt
= int{0 to pi/2} (cos x) / [(cos x) + (sin x)] dx

and int{0 to pi/2} [sin x + cos x] / [sin x + cos x] = pi/2

so, int{0 to pi/2} (sin x) / [(sin x) + (cos x)] dx = pi/4

Fine.

quasi
.