1. Simplify the integrand

We need to evaluate

$$I=\int_0^{\pi/2} \frac{(\sin x+\cos x)^2}{\sqrt{1+\sin 2x}}\,dx.$$

Now,

$$(\sin x+\cos x)^2=\sin^2 x+\cos^2 x+2\sin x\cos x=1+\sin 2x.$$

So the integrand becomes

$$\frac{1+\sin 2x}{\sqrt{1+\sin 2x}}=\sqrt{1+\sin 2x}.$$

Hence,

$$I=\int_0^{\pi/2} \sqrt{1+\sin 2x}\,dx.$$

2. Rewrite $1+\sin 2x$

Using

$$\sin 2x=2\sin x\cos x,$$

we get

$$1+\sin 2x=1+2\sin x\cos x=(\sin x+\cos x)^2.$$

Therefore,

$$\sqrt{1+\sin 2x}=|\sin x+\cos x|.$$

For $x\in[0,\pi/2]$, both $\sin x\ge 0$ and $\cos x\ge 0$, so

$$|\sin x+\cos x|=\sin x+\cos x.$$

Thus,

$$I=\int_0^{\pi/2} (\sin x+\cos x)\,dx.$$

3. Integrate

$$I=\int_0^{\pi/2} \sin x\,dx+\int_0^{\pi/2} \cos x\,dx.$$

Compute each term:

$$\int_0^{\pi/2} \sin x\,dx=[-\cos x]_0^{\pi/2}=0-(-1)=1,$$ $$\int_0^{\pi/2} \cos x\,dx=[\sin x]_0^{\pi/2}=1-0=1.$$

So,

$$I=1+1=2.$$

4. Check options

• A: $3$ ❌
• B: $1$ ❌
• C: $2$ ✅
• D: $0$ ❌

Therefore, the correct answer is Option C.