Question

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

• A. $4\sqrt{2}$
• B. $2\sqrt{2}$
• C. $4\sqrt{3}$
• D. $3\sqrt{2}$

Answer 1

The correct option is B $2\sqrt{2}$

${\int }_0^{\pi }|cosx-sinx|dx$
$=\sqrt{2}{\int }_0^{\pi }\left|sin(\frac{\pi }{4}-x)\right|dx$

$=\sqrt{2}{\int }_0^{\frac{\pi }{4}}\left|sin(\frac{\pi }{4}-x)\right|dx+\sqrt{2}{\int }_{\frac{\pi }{4}}^{\pi }\left|\sin (\frac{\pi }{4}-x)\right|dx$

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

$=\sqrt{2}[{\left[cos(x-\frac{\pi }{4})\right]}_0^{\pi /4}-{\left[cos(x-\frac{\pi }{4})\right]}_{\pi /4}^{\pi }]$

$=\sqrt{2}[+(1-\frac{1}{\sqrt{2}})-(\frac{-1}{\sqrt{2}}-1)]$

$=\sqrt{2}\left[2\right]=2\sqrt{2}$