A calculus problem by Akshat Sharda

$\begin{aligned} I & = \int_0^{2 \pi} \bigg|\sqrt{15}\sin x + \cos x \bigg| dx \\ & = \int_0^{2 \pi} \bigg|4\sin {\color{#3D99F6}(x + \phi)} \bigg| dx & \small \color{#3D99F6} \text{where }\phi = \tan^{-1} \frac 1{\sqrt{15}} \\ & = \int_\phi^{2 \pi+\phi} \bigg|4\sin {\color{#3D99F6}\theta} \bigg| d\theta & \small \color{#3D99F6} \text{Let }\theta = x + \phi \implies d \theta = dx \\ & = 4 \int_{\color{#3D99F6}0}^{\color{#3D99F6}2\pi} |\sin \theta| \ d\theta & \small \color{#3D99F6} \text{See note.} \\ & = {\color{#D61F06}8} \int_{\color{#D61F06}0}^{\color{#D61F06}\pi} \sin \theta \ d\theta \\ & = 8 \bigg[-\cos \theta \bigg]_0^\pi \\ & = \boxed{16} \end{aligned}$

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Note:

$\begin{aligned} \int_\phi^{2\pi + \phi} |\sin \theta| \ d\theta & = \int_\phi^\pi \sin \theta \ d \theta - \int_\pi^{2\pi} \sin \theta \ d \theta + \int_{2\pi}^{2\pi+\phi} \sin \theta \ d \theta \\ & = {\color{#3D99F6} \int_\phi^\pi \sin \theta \ d \theta} - \int_\pi^{2\pi} \sin \theta \ d \theta + {\color{#3D99F6} \int_{\color{#D61F06}0}^{\color{#D61F06}\phi} \sin \theta \ d \theta} \\ & = {\color{#3D99F6} \int_{\color{#D61F06}0}^\pi \sin \theta \ d \theta} - \int_\pi^{2\pi} \sin \theta \ d \theta \\ & = \int_0^{2\pi} |\sin \theta| \ d \theta \end{aligned}$