Question

evaluate the integration of under root 1-sin2x dx​

Answer 1

√1-sin 2x =

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Answer 2

Answer:

$\sqrt{1 - \sin2x} \\ \\ = > \sqrt{ { \sin }^{2} x + { \cos }^{2}x - 2 \sin(x) \cos(x) } \\ \\ = > \sqrt{ {( \sin(x) - \cos(x) ) }^{2} } \\ \\ = > \sin(x) - \cos(x)$

Now, integration of sinx-cosx = -cosx-sinx+c

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