Integral finding without "traditional" ways calculating

Question

I want to find the: $$\int_0^\frac{\pi}{2}\frac{\sin(x)}{\sin(x)+\cos(x)}dx$$ but I think that with the "traditional" ways would be very complicated, so I have prove that: $$\int_0^af(x)dx=\int_0^af(a-x)dx$$ Now I have that: $$\int_0^\frac{\pi}{2}\frac{\sin(x)}{\sin(x)+\cos(x)}dx=\int_0^\frac{\pi}{2}\frac{\cos(\frac{\pi}{2}-x)}{\cos(\frac{\pi}{2}-x)+\cos(x)}dx$$ but I don't know how to continue. Any ideas?

Answer 1

Hint:

$$\int_0^{\frac{\pi}{2}}\frac{\sin(x)}{\sin(x)+\cos(x)}\, dx + \int_0^{\frac{\pi}{2}}\frac{\cos(x)}{\sin(x)+\cos(x)}\, dx=\frac{\pi}{2}$$

You just have to figure out another relationship between $\int_0^{\frac{\pi}{2}}\frac{\sin(x)}{\sin(x)+\cos(x)}\, dx $ and $\int_0^{\frac{\pi}{2}}\frac{\cos(x)}{\sin(x)+\cos(x)}\, dx $

Edit:

$$\color{blue}{\int_0^\frac{\pi}{2}\frac{\sin(x)}{\sin(x)+\cos(x)}\, dx}= \int_0^\frac{\pi}{2}\frac{\sin\left(\frac{\pi}{2}-x\right)}{\sin\left(\frac{\pi}{2}-x\right)+\cos\left(\frac{\pi}{2}-x\right)}\, dx= \color{green}{\int_0^\frac{\pi}{2}\frac{\cos(x)}{\sin(x)+\cos(x)}\, dx}\tag{1}$$

$$\color{blue}{\int_0^\frac{\pi}{2}\frac{\sin(x)}{\sin(x)+\cos(x)}\, dx} +\color{green}{\int_0^\frac{\pi}{2}\frac{\cos(x)}{\sin(x)+\cos(x)}\, dx}=\int_0^\frac{\pi}{2} \frac{\sin(x)+\cos(x)}{\sin(x)+\cos(x)}\, dx=\int_0^\frac{\pi}{2} \, dx = \frac{\pi}{2} \tag{2}$$

Substitute $(1)$ into $(2)$,

$$\color{blue}{\int_0^\frac{\pi}{2}\frac{\sin(x)}{\sin(x)+\cos(x)}\, dx} + \color{blue}{\int_0^\frac{\pi}{2}\frac{\sin(x)}{\sin(x)+\cos(x)}\, dx} = \frac{\pi}{2}$$

That is $$2I=\frac{\pi}{2}$$

Answer 2

While using the result $$\int_a^b f(x) \, dx = \int_0^a f(a - x) \, dx,$$ to evaluate the integral $$\int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} \, dx,$$ is by far the best approach to use in this instance, it is not too hard to evaluate this particular integral using what you refer to as the “traditional” approach.

Noting that $\sin x + \cos x = \sqrt{2} \cos (x - \pi/4)$, the integral can be written as $$I = \frac{1}{\sqrt{2}} \int_0^{\pi/2} \frac{\sin x}{\cos (x - \pi/4)} \, dx.$$ Enforcing a substitution of $x \mapsto x + \pi/4$ leads to \begin{align*} I &= \frac{1}{\sqrt{2}} \int_{-\pi/4}^{\pi/4} \frac{\sin (x + \pi/4)}{\cos x} \, dx\\ &= \int_{-\pi/4}^{\pi/4} \frac{\sin x + \cos x}{\sin x} \, dx\\ &= \int_{-\pi/4}^{\pi/4} (\tan x + 1) \, dx\\ &= \Big{[} x - \ln (\cos x) \Big{]}_{-\pi/4}^{\pi/4}\\ &= \frac{\pi}{2}, \end{align*} as expected.