无需传统方法求解定积分∫₀^(π/2)sinx/(sinx+cosx)dx的思路咨询

Hey there! Nice work recognizing that symmetry formula—it’s a total game-changer for integrals like this, and you’re already on the right track. Let’s break down the next steps to wrap this up:

• First, let’s assign a name to your original integral to make things cleaner. Let’s say:
  $$I = \int_0^\frac{\pi}{2}\frac{\sin(x)}{\sin(x)+\cos(x)}dx$$

• You correctly applied the identity ( \int_0^{af(x)dx=\int_0}af(a-x)dx ), but let’s tweak that result slightly to make it easier to work with. When we substitute ( x \to \frac{\pi}{2} - x ):

  • ( \sin\left(\frac{\pi}{2} - x\right) = \cos(x) )
  • ( \cos\left(\frac{\pi}{2} - x\right) = \sin(x) )
    So this transforms your integral into:
    $$I = \int_0^\frac{\pi}{2}\frac{\cos(x)}{\cos(x)+\sin(x)}dx$$

• Now, here’s the key trick: add the original expression for ( I ) to this new version of ( I ):
  $$I + I = \int_0^\frac{\pi}{2}\frac{\sin(x)}{\sin(x)+\cos(x)}dx + \int_0^\frac{\pi}{2}\frac{\cos(x)}{\sin(x)+\cos(x)}dx$$

• Combine the two integrals (since they have the same limits and denominator):
  $$2I = \int_0^\frac{\pi}{2}\frac{\sin(x) + \cos(x)}{\sin(x) + \cos(x)}dx$$

• The numerator and denominator cancel out, leaving a super simple integral:
  $$2I = \int_0^\frac{\pi}{2}1dx$$

• Calculate that basic integral:
  $$\int_0^\frac{\pi}{2}1dx = \frac{\pi}{2}$$

• Finally, solve for ( I ):
  $$2I = \frac{\pi}{2} \implies I = \frac{\pi}{4}$$

That’s it! This symmetry method avoids all the messy trigonometric substitutions or integration by parts you’d have to do with traditional methods. It’s a go-to technique for integrals over symmetric intervals involving trig functions.

内容的提问来源于stack exchange，提问作者Leos Kotrop