Question

The integral \(\displaystyle\int_0^{\pi}\sqrt{1+4\sin^2\frac x2-4\sin\frac x2}\,dx\) equals :

• A. \(4\sqrt3-4\)
• B. \(4\sqrt3-4-\frac{\pi}{3}\)
• C. \(\pi-4\)
• D. \(\frac{2\pi}{3}-4-4\sqrt3\)

Hint:

The radicand is \(\left(1-2\sin\frac x2\right)^2\); split the integral where \(\sin\frac x2=\frac12\), i.e. \(x=\frac\pi3\).

Answer 1

The Correct Option is B

\(1+4\sin^2\frac x2-4\sin\frac x2=\left(1-2\sin\frac x2\right)^2\), so the integrand is \(\left|1-2\sin\frac x2\right|\).
\(1-2\sin\frac x2\ge0\) when \(\sin\frac x2\le\frac12\), i.e. \(0\le x\le\frac\pi3\); it is negative on \(\left(\frac\pi3,\pi\right]\).
Put \(u=\frac x2\), \(dx=2du\): \(I=2\left[\int_0^{\pi/6}(1-2\sin u)\,du+\int_{\pi/6}^{\pi/2}(2\sin u-1)\,du\right]\).
First: \(\left[u+2\cos u\right]_0^{\pi/6}=\frac\pi6+\sqrt3-2\). Second: \(\left[-2\cos u-u\right]_{\pi/6}^{\pi/2}=-\frac\pi2+\sqrt3+\frac\pi6=\sqrt3-\frac\pi3\).
Sum \(=2\sqrt3-2-\frac\pi6\); \(I=4\sqrt3-4-\frac\pi3\).
Answer: (2) \(4\sqrt3-4-\frac{\pi}{3}\)