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

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  ,