Cómo resolver esta ecuación de valor absoluto

Question

Tengo la siguiente integral$$\int_0^{\pi/2} |\sin x-\cos x|dx. $$It's a simple integral but when I try to solve the module I get stuck. I took $ \ sin x- \ cos x> 0$ and squaring this I found that $ \ sin 2x <1$. When I apply $ \ arcsin$ it would mean $$2x<\frac\pi2 \implies x<\frac\pi4$$ but the interval is wrong from the one in my book. When I apply $ \ arcsin $ ¿cambia el signo? ¿Por qué?

Answer 1

Tenga en cuenta que $\sin x-\cos x=\cos x(\tan x-1)$.

$\sin x\le\cos x$ cuando $x\in[0,\frac{\pi}{4}]$.

$\sin x\ge\cos x$ cuando $x\in[\frac{\pi}{4},\frac{\pi}{2}]$.

ps

Answer 2

Sugerencia sobre cómo determinar el signo de$\sin x - \cos x$:

ps

Answer 3

$$ \begin{align} \int_0^{\pi/2}|\sin(x)-\cos(x)|\,\mathrm{d}x &=\int_0^{\pi/4}(\cos(x)-\sin(x))\,\mathrm{d}x +\int_{\pi/4}^{\pi/2}(\sin(x)-\cos(x))\,\mathrm{d}x\tag{1}\\ &=2\int_0^{\pi/4}(\cos(x)-\sin(x))\,\mathrm{d}x\tag{2}\\ &=2\left[\vphantom{\int}\sin(x)+\cos(x)\right]_0^{\pi/4}\tag{3}\\[6pt] &=2\sqrt2-2\tag{4} \end {align} $$ Explicación:
$(1)$: $\cos(x)\ge\sin(x)$ incesantemente $\left[0,\frac\pi4\right]$
$\sin(x)\ge\cos(x)$:$\left[\frac\pi4,\frac\pi2\right]$ y$(2)$
$\cos(x)=\sin\left(\frac\pi2-x\right)$: integrar
$\sin(x)=\cos\left(\frac\pi2-x\right)$: evaluar

Answer 4

La función$\;f(x)=\lvert \sin x-\cos x\rvert\;$ tiene una simetría wrt$\frac\pi 4$ desde$\;f\bigl(\frac \pi2-x\bigr)=f(x)$, de ahí$$\int_0^{\tfrac\pi 2} \lvert\sin x-\cos x\rvert \,\mathrm dx=2\int_0^{\tfrac\pi 4} (\cos x-\sin x) \,\mathrm dx=2(\sin x+\cos x)\biggl\rvert_0^{\tfrac\pi4}=2(\sqrt2-1).$ $

Answer 5

\begin{align} \int_0^{\pi/2} |\sin(x)-\cos(x)|dx &= \int_0^{\pi/4} |\sin(x)-\cos(x)|dx+\int^0_{\pi/4} |\sin(x)-\cos(x)|dx\\ &=-\int_0^{\pi/4} \sin(x)-\cos(x)dx+\int_{\pi/4}^{\pi/2} \sin(x)-\cos(x)dx\\ &=.. \end{align}