Question

Solve:

$\underset{0}{\overset{\pi /2}{\int }}\left|\sin x-\cos x\right|dx=$

• A. $0$
• B. $2-2\sqrt{2}$
• C. $2\sqrt{2}$
• D. $2(\sqrt{2}+1)$

Answer 1

Answer: (B)

$2-2\sqrt{2}$

Given that:

${\int }_0^{\frac{\pi }{2}}|\sin x-\cos x|dx$

Since

$\cos x\ge \sin x$, $x\in \left[0,\frac{\pi }{4}\right]$

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

So,

$={\int }_0^{\frac{\pi }{4}}-(\sin x-\cos x)dx+{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\sin x-\cos xdx$

$={[\cos x+\sin x]}_0^{\frac{\pi }{4}}+{[-\cos x-\sin x]}_{\frac{\pi }{4}}^{\frac{\pi }{2}}$

$=\cos \left(0\right)+\sin \left(0\right)-\cos \left(\frac{\pi }{4}\right)-\sin \left(\frac{\pi }{4}\right)-\cos \left(\frac{\pi }{4}\right)-\sin \left(\frac{\pi }{4}\right)+\cos \left(\frac{\pi }{2}\right)+\sin \left(\frac{\pi }{2}\right)$

$=1+0-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}+1$

$=2-2\sqrt{2}$