(int\frac{sin x+cos x}{sqrt{1 + sin 2x}}dx)

Explanation:

Step1: Simplify the denominator

Since $\sin2x = 2\sin x\cos x$, then $1+\sin2x=\sin^{2}x + 2\sin x\cos x+\cos^{2}x=(\sin x+\cos x)^{2}$. So, $\sqrt{1 + \sin2x}=\vert\sin x+\cos x\vert$.

Step2: Rewrite the integral

The integral $\int\frac{\sin x+\cos x}{\sqrt{1 + \sin2x}}dx=\int\frac{\sin x+\cos x}{\vert\sin x+\cos x\vert}dx$. Case 1: When $\sin x+\cos x>0$, i.e., $\sqrt{2}\sin(x + \frac{\pi}{4})>0$, the integral becomes $\int 1dx=x + C_1$. Case 2: When $\sin x+\cos x<0$, i.e., $\sqrt{2}\sin(x+\frac{\pi}{4})<0$, the integral becomes $\int(- 1)dx=-x + C_2$.

Answer:

$\begin{cases}x + C_1, &\sin x+\cos x>0\-x + C_2, &\sin x+\cos x<0\end{cases}$