Question

If $[.]$denotes G.I F, ${\int }_0^{2\pi }[|sinx|-|cosx|]dx=$

• A. $-2\pi$
• B. $-4\pi$
• C. $-\pi$
• D. 0

Answer 1

The correct option is C

$-\pi$

$|\sin x|-|\cos x|\in [-1,1]\therefore {\int }_0^{2\pi }[|sinx|-|cosx|]dx=2{\int }_0^{\pi }[|sinx|-|cosx|]dx=2\left({\int }_0^{\frac{\pi }{4}}[|sinx|-|cosx|]dx+{\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}[|sinx|-|cosx|]dx+{\int }_{\frac{3\pi }{4}}^{\pi }[|sinx|-|cosx|]dx\right)$$=2\left({\int }_0^{\frac{\pi }{4}}(-1)dx+{\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}0dx+{\int }_{\frac{3\pi }{4}}^{\pi }(-1)dx\right)=-\pi$