Test series

Question

$\int_{0}^{2\pi } ( \sqrt{1 - sin 2x }) dx$

= $\int_{0}^{2\pi } ( \sqrt{sin^{2}x + cos^{2}x - 2sinxcosx }) dx$

=$\int_{0}^{2\pi } | sin x - cos x | dx$

after this how to break into interval  please help

Comments

$\sqrt{32}=4\sqrt{2} \ ?$

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$\Large\int_0^{\frac{\pi}{4}} (cosx-sinx) \ dx \ +\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (sinx-cosx) \ dx +\int_{\frac{5\pi}{4}}^{2\pi} (cosx-sinx) \ dx$

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 Inspiron  yes so ?

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Inspiron thnks how you get that  tell method  ?

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Graphs.

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Inspiron  is there any other method , you know  ,if function is complex then in exam if its graph is complex then my entire time will be over in drawing graph only

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https://math.stackexchange.com/questions/2627776/definite-integral-sinx-cosxdx#2627776

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thanku  bhai Tuhin Dutta edit your post there asking how they are finding interval  ,