int frac{cos x - sin x}{1 + sin 2x} , dx = | vedclass

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(A) We have the integral $I = \int \frac{\cos x - \sin x}{1 + \sin 2x} \, dx$.Since $1 + \sin 2x = \sin^2 x + \cos^2 x + 2 \sin x \cos x = (\sin x + \

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$-\frac{1}{\cos x + \sin x} + c$

Vedclass · Answer

(A) We have the integral $I = \int \frac{\cos x - \sin x}{1 + \sin 2x} \, dx$.Since $1 + \sin 2x = \sin^2 x + \cos^2 x + 2 \sin x \cos x = (\sin x + \cos x)^2$,the integral becomes:$I = \int \frac{\cos x - \sin x}{(\sin x + \cos x)^2} \, dx$.Let $t = \sin x + \cos x$.Then $dt = (\cos x - \sin x) \, dx$.Substituting these into the integral,we get:$I = \int \frac{1}{t^2} \, dt = \int t^{-2} \, dt = \frac{t^{-1}}{-1} + c = -\frac{1}{t} + c$.Substituting back $t = \sin x + \cos x$,we get:$I = -\frac{1}{\sin x + \cos x} + c$.

$\int \frac{\cos x - \sin x}{1 + \sin 2x} \, dx = $

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