integration making cases

Question

$$\int_{0}^{2\pi}|\sin x - \cos x|\,dx.$$

now we will try to make cases , what i did i put sin x - cos x = 0

this gives tan x = 0 , but iam stuck here

I applied above logic by taking below in consideration

$$\int_{0}^{6}|x - 3|\,dx.$$ x -3 = 0 , x =3

$$\int_{0}^{6}|x - 3|\,dx.$$ = $$\int_{0}^{3}3-x\,dx.$$ + $$\int_{3}^{6}x - 3\,dx.$$

Answer 1

I can't understand what you tried to do, but

$$x\in\left[0,\,2\pi\right]\;,\;\;\cos x-\sin x\ge0\iff x\in\left[0,\,\frac\pi4\right]\cup\left[\frac{5\pi}4,\,2\pi\right]$$

so

$$\int_0^{2\pi}|\cos x-\sin x|dx=\int_0^{\pi/4}(\cos x-\sin x)dx+\int_{5\pi/4}^{2\pi}(\cos x-\sin x)dx+$$

$$+\int_{\pi/4}^{5\pi/4}(\sin x-\cos x)dx=4\sqrt2$$

Answer 2

Begin with simplifying the integrand: $$\sin x-\cos x=\sqrt{2\rule{0pt}{2ex}}\sin\Bigl(x-\dfrac\pi4\Bigr),$$ so that by substitution ($u=x-\frac\pi4$): $$\int_{0}^{2\pi}|\sin x - \cos x|\,\mathrm dx=\sqrt{2\rule{0pt}{2ex}}\int_{-\tfrac\pi4}^{\tfrac{7\pi}4}|\sin u|\,\mathrm du=\sqrt{2\rule{0pt}{2ex}}\int_{0}^{2\pi}|\sin u|\,\mathrm du=2\sqrt{2\rule{0pt}{2ex}}\int_{0}^{\pi}\sin u\,\mathrm du.$$

Answer 3

You do not need to separate cases. For any $C>0$ and any $\theta\in\mathbb{R}$ we have $$ \int_{0}^{2\pi}\left|C\sin(x+\theta)\right|\,dx = 4C $$ since both $\left|\sin x\right|$ and $\left|\sin (x+\theta)\right|$ are $\pi$-periodic and their integral over a period equal $2$.
Your problem can be solved by considering $C=\sqrt{2}$ and $\theta=-\frac{\pi}{4}$.