The integral $$\int_0^{\pi} \sqrt{1 + 4\sin^2\frac{x}{2} - 4\sin\frac{x}{2}} \, dx$$ equals:

We begin with the definite integral

$$I=\int_{0}^{\pi}\sqrt{\,1+4\sin^2\frac{x}{2}-4\sin\frac{x}{2}\,}\,dx.$$

First we simplify the quantity inside the square root. We have the algebraic expression

$$1+4\sin^2\frac{x}{2}-4\sin\frac{x}{2}.$$

Because the highest power of $$\sin\dfrac{x}{2}$$ is two, it is natural to complete the square. We write

$$4\sin^2\frac{x}{2}-4\sin\frac{x}{2}=4\Bigl(\sin^2\frac{x}{2}-\sin\frac{x}{2}\Bigr).$$

Now we recall the standard identity for completing a square: for any real number $$a$$,

$$a^2-a=\Bigl(a-\tfrac12\Bigr)^2-\tfrac14.$$

Using $$a=\sin\dfrac{x}{2}$$, we obtain

$$\sin^2\frac{x}{2}-\sin\frac{x}{2}=\Bigl(\sin\frac{x}{2}-\tfrac12\Bigr)^2-\tfrac14.$$

Multiplying by 4 gives

$$4\Bigl(\sin^2\frac{x}{2}-\sin\frac{x}{2}\Bigr) =4\Bigl(\sin\frac{x}{2}-\tfrac12\Bigr)^2-1.$$ Adding the remaining constant 1 from the original radicand yields

$$1+4\sin^2\frac{x}{2}-4\sin\frac{x}{2} =4\Bigl(\sin\frac{x}{2}-\tfrac12\Bigr)^2.$$

Therefore the square root simplifies beautifully:

$$\sqrt{1+4\sin^2\frac{x}{2}-4\sin\frac{x}{2}} =\sqrt{4}\,\Bigl|\sin\frac{x}{2}-\tfrac12\Bigr| =2\Bigl|\sin\frac{x}{2}-\tfrac12\Bigr|.$$

So the integral becomes

$$I=2\int_{0}^{\pi}\Bigl|\sin\frac{x}{2}-\tfrac12\Bigr|\,dx.$$

To remove the fractional argument inside the sine function we make the substitution

$$u=\frac{x}{2}\;\;\Longrightarrow\;\;x=2u,\qquad dx=2\,du.$$

When $$x=0$$, $$u=0$$, and when $$x=\pi$$, $$u=\dfrac{\pi}{2}$$. Substituting gives

$$I=2\int_{0}^{\pi}\Bigl|\sin\frac{x}{2}-\tfrac12\Bigr|\,dx =2\int_{0}^{\pi}\Bigl|\sin\frac{x}{2}-\tfrac12\Bigr|\,(dx) =2\int_{0}^{\pi}\Bigl|\sin\frac{x}{2}-\tfrac12\Bigr|\,(2\,du)$$

$$\;\;=4\int_{0}^{\pi/2}\Bigl|\sin u-\tfrac12\Bigr|\,du.$$

Next we analyse the absolute value. Inside the interval $$0\le u\le\dfrac{\pi}{2}$$ the function $$\sin u$$ increases monotonically from 0 to 1. The point where $$\sin u=\dfrac12$$ is

$$u=\frac{\pi}{6},\quad\text{because}\quad\sin\frac{\pi}{6}=\frac12.$$

Hence

• On $$0\le u\le\dfrac{\pi}{6}$$ we have $$\sin u\le\dfrac12$$, so $$\bigl|\sin u-\tfrac12\bigr|=\tfrac12-\sin u.$$
• On $$\dfrac{\pi}{6}\le u\le\dfrac{\pi}{2}$$ we have $$\sin u\ge\dfrac12$$, so $$\bigl|\sin u-\tfrac12\bigr|=\sin u-\tfrac12.$$

Splitting the integral at $$u=\dfrac{\pi}{6}$$ we get

$$I=4\Bigl[\;\int_{0}^{\pi/6}\Bigl(\tfrac12-\sin u\Bigr)\,du +\int_{\pi/6}^{\pi/2}\Bigl(\sin u-\tfrac12\Bigr)\,du\Bigr].$$

We now evaluate each integral separately.

First part:

$$\int\Bigl(\tfrac12-\sin u\Bigr)\,du =\int\tfrac12\,du-\int\sin u\,du =\frac{u}{2}+\cos u.$$

Thus

$$\int_{0}^{\pi/6}\Bigl(\tfrac12-\sin u\Bigr)\,du =\Bigl[\frac{u}{2}+\cos u\Bigr]_{0}^{\pi/6} =\frac{\pi}{12}+\cos\frac{\pi}{6}-\bigl(0+\cos0\bigr) =\frac{\pi}{12}+\frac{\sqrt3}{2}-1.$$

Second part:

$$\int\Bigl(\sin u-\tfrac12\Bigr)\,du =\int\sin u\,du-\int\tfrac12\,du =(-\cos u)-\frac{u}{2}.$$

Hence

$$\int_{\pi/6}^{\pi/2}\Bigl(\sin u-\tfrac12\Bigr)\,du =\Bigl[-\cos u-\frac{u}{2}\Bigr]_{\pi/6}^{\pi/2} =\Bigl(-\cos\frac{\pi}{2}-\frac{\pi}{4}\Bigr) -\Bigl(-\cos\frac{\pi}{6}-\frac{\pi}{12}\Bigr)$$

$$=-0-\frac{\pi}{4}+\frac{\sqrt3}{2}+\frac{\pi}{12} =-\frac{\pi}{6}+\frac{\sqrt3}{2}.$$

Now we add the two pieces:

$$\frac{\pi}{12}+\frac{\sqrt3}{2}-1 +\Bigl(-\frac{\pi}{6}+\frac{\sqrt3}{2}\Bigr) =\Bigl(\frac{\pi}{12}-\frac{\pi}{6}\Bigr) +\Bigl(\frac{\sqrt3}{2}+\frac{\sqrt3}{2}\Bigr)-1 =-\frac{\pi}{12}+\sqrt3-1.$$

Finally we multiply by the outside factor 4:

$$I=4\Bigl(-\frac{\pi}{12}+\sqrt3-1\Bigr) =-\frac{\pi}{3}+4\sqrt3-4 =4\sqrt3-4-\frac{\pi}{3}.$$

Therefore the value of the original integral is

$$\boxed{\,4\sqrt3-4-\dfrac{\pi}{3}\,}.$$

Comparing with the given options, this matches Option B.

Hence, the correct answer is Option B.