Answer :

We know that

1 + sin2x = sin2x + cos2x + 2sinxcosx


= (sin x + cos x)2




Let, sin x + cos x = t


Differentiating both sides with respect to x



-dt = (sin x – cos x)dx



Use formula


y = -log t + c


Again, put t = sin x + cos x


y = -log(sin x + cos x) + c



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