Indian Statistical Institute B.Math & B.Stat : Integration

Evaluate $$\int_{0}^{2 \pi} |1+2 \sin x| dx$$ when $x \in [0,\pi] \quad \sin x \geq 0 \quad \implies |1+2 \sin x| = 1+2 \sin x$ Now note that when $-1 \leq \sin x < - \frac{1}{2} \quad \implies -2 \leq 2 \sin x < -1 \quad \implies -1 \leq 1+ 2 \sin x < 0 \quad \implies |1+2 \sin x| = -(1+2 \sin x)$ In the $third-quadrant$ sine function continuously decreases (strictly) from $0$ to $-1$ and $\sin \frac{7 \pi}{6} = - \frac{1}{2}$ and at $x=\frac{3 \pi}{2}$ it takes the value $-1$. After that int the $fourth-quadrant$ sine function continuously increases (strictly) from $-1$ to $0$ and at $x=\frac{11 \pi}{6}$ it takes the value $- \frac{1}{2}$. So we conclude that in the $third$ and the $fourth-quadrant,$ $-1 \leq \sin x < - \frac{1}{2} \quad when \quad x \in [\frac{7 \pi}{6},\frac{11 \pi}{6}]$. So, $$\int_{0}^{2 \pi} |1+2 \sin x| dx\ = \int_{0}^{\frac{7 \pi}{6}} (1+2 \sin x) dx+ \int_{\frac{11 \pi}{6}}^{2 \pi} (1+2 \sin x) dx- \int_{\frac{7 \pi}{6}}^{\frac{11 \pi}{6}} (1+2 \sin x) dx$$ $$= [x-2 \cos x]_{0}^{\frac{7 \pi}{6}}+ [x-2 \cos x]_{\frac{11 \pi}{6}}^{2 \pi}- [x-2 \cos x]_{\frac{7 \pi}{6}}^{\frac{11 \pi}{6}}$$ $$=\frac{7 \pi}{6}+ \sqrt{3}+2 + 2\pi -2 - \frac{11 \pi}{6} + \sqrt{3}- \frac{11 \pi}{6} + \sqrt{3} + \frac{7 \pi}{6} + \sqrt{3}$$ $$= \frac{2\pi}{3} + 4 \sqrt{3}$$