1. Write the integrand in a simpler form

We need to evaluate $I=\int_0^\pi \sqrt{1+4\sin^2\frac x2-4\sin\frac x2}\,dx.$

Inside the square root, $1+4\sin^2\frac x2-4\sin\frac x2=(2\sin\tfrac x2-1)^2.$

Hence $\sqrt{1+4\sin^2\frac x2-4\sin\frac x2}=|2\sin\tfrac x2-1|.$

So $I=\int_0^\pi |2\sin\tfrac x2-1|\,dx.$

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2. Determine where the expression changes sign

We solve $2\sin\frac x2-1=0 \quad\Rightarrow\quad \sin\frac x2=\frac12.$

Since $x\in[0,\pi]$, we have $\frac x2\in[0,\frac\pi2]$. Thus $\frac x2=\frac\pi6 \Rightarrow x=\frac\pi3.$

Now:

• for $0\le x<\frac\pi3$, $2\sin\frac x2-1<0$
• for $\frac\pi3\le x\le \pi$, $2\sin\frac x2-1\ge 0$

Therefore, $I=\int_0^{\pi/3} \left(1-2\sin\frac x2\right)dx+\int_{\pi/3}^{\pi}\left(2\sin\frac x2-1\right)dx.$

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3. Evaluate the first integral

$I_1=\int_0^{\pi/3} \left(1-2\sin\frac x2\right)dx.$

Now, $\int 2\sin\frac x2\,dx=-4\cos\frac x2,$ so $\int \left(1-2\sin\frac x2\right)dx=x+4\cos\frac x2.$

Thus $I_1=\left[x+4\cos\frac x2\right]_0^{\pi/3}$ $=\left(\frac\pi3+4\cos\frac\pi6\right)-\left(0+4\cos0\right)$ $=\frac\pi3+4\cdot\frac{\sqrt3}{2}-4$ $=\frac\pi3+2\sqrt3-4.$

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4. Evaluate the second integral

$I_2=\int_{\pi/3}^{\pi}\left(2\sin\frac x2-1\right)dx.$

An antiderivative is $\int \left(2\sin\frac x2-1\right)dx=-4\cos\frac x2-x.$

So $I_2=\left[-4\cos\frac x2-x\right]_{\pi/3}^{\pi}$ $=\left(-4\cos\frac\pi2-\pi\right)-\left(-4\cos\frac\pi6-\frac\pi3\right)$ $=(0-\pi)-\left(-2\sqrt3-\frac\pi3\right)$ $=2\sqrt3-\frac{2\pi}{3}.$

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5. Add the two parts

$I=I_1+I_2$ $=\left(\frac\pi3+2\sqrt3-4\right)+\left(2\sqrt3-\frac{2\pi}{3}\right)$ $=4\sqrt3-4-\frac\pi3.$

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6. Match with the options

$I=4\sqrt3-4-\frac\pi3,$ which is Option B.

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7. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.