JEE Main 2014 · 6 Apr · Shift 1, Mathematics question 75

The integrand is $\left|1-2\sin\frac x2\right|$, which changes sign at $x=\frac\pi3$. With $F(x)=x+4\cos\frac x2$, an antiderivative of $1-2\sin\frac x2$:
$$\int_0^{\pi/3}=F\left(\tfrac\pi3\right)-F(0)=\tfrac\pi3+2\sqrt3-4,\qquad\int_{\pi/3}^{\pi}=-\left[F(\pi)-F\left(\tfrac\pi3\right)\right]=2\sqrt3-\tfrac{2\pi}3.$$
Total: $4\sqrt3-4-\dfrac\pi3$.