Đặt \(t = \sqrt {1 - 2\alpha \cos x + {\alpha ^2}} \Rightarrow {t^2} = 1 - 2\alpha \cos x + {\alpha ^2} \Rightarrow \frac{t}{\alpha }dt = \sin x{\rm{d}}x\) Đổi cận: \(\left\{ \begin{array}{l} x = 0 \Rightarrow t = \sqrt {1 - 2\alpha + {\alpha ^2}} = \left| {1 - \alpha } \right| = \alpha - 1\,(\alpha > 1)\\ x = 0 \Rightarrow t = \sqrt {1 + 2\alpha + {\alpha ^2}} = \left| {1 + \alpha } \right| = \alpha + 1\,(\alpha > 1) \end{array} \right.\)
Vậy: \(I = \frac{1}{\alpha }\int\limits_{\alpha - 1}^{\alpha + 1} {\frac{{tdt}}{t}} = \frac{1}{\alpha }.\left. t \right|_{\alpha - 1}^{\alpha + 1} = \frac{2}{\alpha }.\)
