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### Answers (1)

Answer: $-\log |\sin x+\cos x|+c$

Hints: You must know about the integral rule of trigonometric functions

Given: $\int \frac{\sin x-\cos x}{\sqrt{1+\sin 2x}}dx$

Solution: $I=\int \frac{\sin x-\cos x}{\sqrt{1+\sin 2x}}dx$

\begin{aligned} &\int \frac{\sin x-\cos x}{\sqrt{\sin ^{2} x+\cos ^{2} x+2 \sin x} \cdot \cos x} d x \\ &\int \frac{\sin x-\cos x}{\sqrt{(\sin x+\cos x)^{2}}} d x \\ &\int \frac{\sin x-\cos x}{(\sin x+\cos x)} d x \end{aligned}

Let $\sin x+\cos x=t$  and differentiate both sides,  $\left [ \frac{d}{dx}\sin x dx=\cos x \: and\: \frac{d}{dx}\cos x=-\sin x \right ]$

$(\cos x-\sin x)dx=dt$

$-(\sin x-\cos x)dx=dt$

$=\int \frac{-dt}{t}$
$=-\log \left | t \right |+c$                                                                  $\left [ \int \frac{1}{x}dx=\log x +c\right ]$

$=-\log |\sin x+\cos x|+c$

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