Question

The integral ${\int }_0^{\pi /2}\left|sinx-cosx\right|dx$ is equal to :

• A. $2\sqrt{2}$
• B. $2\left(\sqrt{2}-1\right)$
• C. $\sqrt{2}+1$
• D. None of these

Answer 1

Answer: (B)

We have $cosx\ge sinxfor0\le x\le \frac{\pi }{4}$
and $sinx\ge cosxfor\frac{\pi }{4}\le x\le \frac{\pi }{2}$
$\therefore {\int }_0^{\frac{\pi }{2}}\left|sinx-cosx\right|dx$
$={\int }_0^{\frac{\pi }{4}}\left(cosx-sinx\right)dx+{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\left(sinx-cosx\right)dx$
$={\left[sinx+cosx\right]}_0^{\frac{\pi }{4}}+{\left[-cosx-sinx\right]}_{\frac{\pi }{4}}^{\frac{\pi }{2}}$
$=\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-0-1\right]-\left[0+1-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right]$
$\sqrt{2}-1-1+\sqrt{2}=2\sqrt{2}-2$