Question

The value of ${\int }_0^{2\pi }|cosx-sinx|dx$, is

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

Answer 1

The correct option is D $4\sqrt{2}$

$I={\int }_0^{2\pi }|cosx-sinx|dx={\int }_0^{\frac{\pi }{4}}|cosx-sinx|dx+{\int }_{\frac{\pi }{4}}^{\frac{5\pi }{4}}|cosx-sinx|dx+{\int }_{\frac{5\pi }{4}}^{2\pi }|cosx-sinx|dx$
For $0<x<\frac{\pi }{4}$ and $\frac{5\pi }{4}<x<2\pi$
$cosx>sinx$ ...(1)
And for $\frac{\pi }{4}<x<\frac{5\pi }{4}$
$cosx<sinx$ ...(2)
Using (1) and (2), we get
$I={\int }_0^{\frac{\pi }{4}}(cosx-sinx)dx-{\int }_{\frac{\pi }{4}}^{\frac{5\pi }{4}}(cosx-sinx)dx+{\int }_{\frac{5\pi }{4}}^{2\pi }(cosx-sinx)dx={[sinx+cosx]}_0^{\frac{\pi }{4}}-{[sinx+cosx]}_{\frac{\pi }{4}}^{\frac{5\pi }{4}}+{[sinx+cosx]}_{\frac{5\pi }{4}}^{2\pi }=(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-0-1)-(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}})+(0+1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})=\sqrt{2}-1+2\sqrt{2}+1-\sqrt{2}=4\sqrt{2}$