int frac{cos x-sin x}{5+sin (2 x)} d x= | vedclass

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(B) Let $I = \int \frac{\cos x-\sin x}{5+\sin 2 x} d x$. We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin x - \cos x)^2$ is not helpful here,but

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$\frac{1}{2} 	an ^{-1}\left[\frac{1}{2}(\sin x+\cos x)ight]+c$

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(B) Let $I = \int \frac{\cos x-\sin x}{5+\sin 2 x} d x$. We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin x - \cos x)^2$ is not helpful here,but $(\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2 \sin x \cos x = 1 + \sin 2x$. Thus,$\sin 2x = (\sin x + \cos x)^2 - 1$. Substituting this into the integral: $I = \int \frac{\cos x-\sin x}{5 + (\sin x + \cos x)^2 - 1} d x = \int \frac{\cos x-\sin x}{4 + (\sin x + \cos x)^2} d x$. Let $t = \sin x + \cos x$. Then $dt = (\cos x - \sin x) dx$. Substituting $t$ and $dt$ into the integral: $I = \int \frac{dt}{4 + t^2} = \frac{1}{2} 	an^{-1}\left(\frac{t}{2}ight) + C$. Substituting back $t = \sin x + \cos x$: $I = \frac{1}{2} 	an^{-1}\left(\frac{\sin x + \cos x}{2}ight) + C$.

$\int \frac{\cos x-\sin x}{5+\sin (2 x)} d x=$

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