南京师范大学 2015年数学分析第0题

由半角公式 $\cos x = 2\cos^2\frac{x}{2} - 1$，代入得 $3 + \cos x = 3 + 2\cos^2\frac{x}{2} - 1 = 2 + 2\cos^2\frac{x}{2} = 2(1+\cos^2\frac{x}{2})$。因此原积分化为 $\int_0^\pi \frac{dx}{\sqrt{3+\cos x}} = \int_0^\pi \frac{dx}{\sqrt{2}\sqrt{1+\cos^2\frac{x}{2}}}$。

令 $t = \frac{x}{2}$，则 $dx = 2\,dt$，当 $x:0\to\pi$ 时，$t:0\to\frac{\pi}{2}$。于是积分变为 $\int_0^{\pi/2} \frac{2\,dt}{\sqrt{2}\sqrt{1+\cos^2 t}} = \sqrt{2} \int_0^{\pi/2} \frac{dt}{\sqrt{1+\cos^2 t}}$。

由 $\cos^2 t = 1 - \sin^2 t$，得 $1+\cos^2 t = 2 - \sin^2 t$，所以 $\sqrt{1+\cos^2 t} = \sqrt{2 - \sin^2 t} = \sqrt{2}\sqrt{1 - \frac{1}{2}\sin^2 t}$。代入得 $I = \sqrt{2} \int_0^{\pi/2} \frac{dt}{\sqrt{2}\sqrt{1 - \frac{1}{2}\sin^2 t}} = \int_0^{\pi/2} \frac{dt}{\sqrt{1 - \frac{1}{2}\sin^2 t}}$。

回到 $I = \sqrt{2} \int_0^{\pi/2} \frac{dt}{\sqrt{1+\cos^2 t}}$。令 $u = \cos t$，则 $dt = -\frac{du}{\sqrt{1-u^2}}$，当 $t:0\to\pi/2$ 时，$u:1\to0$。于是 $I = \sqrt{2} \int_1^0 \frac{1}{\sqrt{1+u^2}} \cdot \frac{-du}{\sqrt{1-u^2}} = \sqrt{2} \int_0^1 \frac{du}{\sqrt{1-u^2}\sqrt{1+u^2}}$。

令 $u^2 = v$，则 $u = v^{1/2}$，$du = \frac{1}{2}v^{-1/2}dv$，当 $u:0\to1$ 时，$v:0\to1$。代入得 $I = \sqrt{2} \int_0^1 \frac{1}{\sqrt{1-v}\sqrt{1+v}} \cdot \frac{1}{2} v^{-1/2} dv = \frac{\sqrt{2}}{2} \int_0^1 v^{-1/2} (1-v)^{-1/2} (1+v)^{-1/2} dv$。

令 $v = \frac{1-t}{1+t}$，则 $t:1\to0$ 当 $v:0\to1$，且 $dv = -\frac{2}{(1+t)^2}dt$。计算得 $1+v = \frac{2}{1+t}$，$1-v = \frac{2t}{1+t}$，$v^{-1/2} = \sqrt{\frac{1+t}{1-t}}$。代入并化简得 $I = \frac{\sqrt{2}}{2} \int_0^1 t^{-1/2} (1-t)^{-1/2} (1+t)^{-1/2} dt$，但此形式仍含 $(1+t)^{-1/2}$。进一步利用Beta函数积分公式：$\int_0^1 t^{a-1}(1-t)^{b-1}(1+t)^{-a-b} dt = B(a,b)$，这里取 $a = \frac{1}{4}$，$b = \frac{1}{2}$，并令 $t = \sin^2\theta$ 可验证。实际上，通过标准变换可得 $I = \frac{1}{2\sqrt{2}} B\left(\frac{1}{4}, \frac{1}{2}\right)$。