积分练习 3

1

$$ I=\int_2^{\infty}\dfrac{e^{-\sqrt{x}}}{(\sqrt{x}-1)^2}\text{d}x $$

【Sol】:
$\begin{aligned} I=&\int_{\sqrt{2}}^{\infty}\dfrac{e^{-x}}{(x-1)^2}(2x\text{d}x)\\
=&2\int_{\sqrt{2}}^{\infty}\dfrac{xe^{-x}}{(x-1)^2}\text{d}x\\
=&2\int_{\sqrt{2}-1}^{\infty}\dfrac{x+1}{x^2}e^{-x-1}\text{d}x\\
=&\dfrac{2}{e}\int_{\sqrt{2}-1}^{\infty}(x+1)e^{-x}\text{d}\left(-\dfrac{1}{x}\right)\\
=&\dfrac{2}{e}\left[-\dfrac{x+1}{x}e^{-x}\bigg|_{\sqrt{2}-1}^{\infty}+\int_{\sqrt{2}-1}^{\infty}\dfrac{1}{x}(-x)e^{-x}\text{d}x\right]\\
=&\dfrac{2}{e}\left[\dfrac{\sqrt{2}}{\sqrt{2}-1}e^{1-\sqrt{2}}-\int_{\sqrt{2}-1}^{\infty}e^{-x}\text{d}x\right]\\
=&\dfrac{2}{e}\left[\sqrt{2}(\sqrt{2}+1)e^{1-\sqrt{2}}-e^{1-\sqrt{2}}\right]\\
=&2(\sqrt{2}+1)e^{-\sqrt{2}} \end{aligned}$

2

$$ I=\int_0^1\dfrac{1}{1+\sqrt{2x\sqrt{1-x^2}}}\text{d}x $$

【Sol】:
利用公式
$$ \int_0^1f(2x\sqrt{1-x^2})\text{d}x=\int_0^1f(1-x^2)\text{d}x $$ 于是有:
$\begin{aligned} I=&\int_0^1\dfrac{1}{1+\sqrt{1-x^2}}\text{d}x\\
=&\int_0^{\frac{\pi}{2}}\dfrac{1}{1+\cos x}\text{d}(\sin x)\\
=&\int_0^{\frac{\pi}{2}}\left[1-\dfrac{1}{1+\cos x}\right]\text{d}x\
=&\dfrac{\pi}{2}-\dfrac{\sin x}{1+\cos x}\bigg|_{0}^{\frac{\pi}{2}}\
=&\dfrac{\pi}{2}-1 \end{aligned}$

3

$$ I=\int_{0}^{2\pi}\dfrac{\sin^2x}{5+4\cos x}\text{d}x=\dfrac{\pi}{4} $$

【Sol】:
被积函数是周期为 $2\pi$ 的偶函数，因此可以将区间移动到对称区间 $[-\pi,; \pi]$ 上:
$$ I=\int_{-\pi}^{\pi}\dfrac{\sin^2x}{5+4\cos x}\text{d}x=2\int_{0}^{\pi}\dfrac{\sin^2x}{5+4\cos x}\text{d}x $$

又余弦函数在 $[0,; \pi]$ 上不对称，因此
$\begin{aligned} I=&2\left[\int_0^{\frac{\pi}{2}}\dfrac{\sin^2x}{5+\cos x}\text{d}x+\int_{\frac{\pi}{2}}^{\pi}\dfrac{\sin^2x}{5+4\cos x}\text{d}x\right]\\
=&2\left[\int_0^{\frac{\pi}{2}}\dfrac{\sin^2x}{5+4\cos x}\text{d}x+\int_{0}^{\frac{\pi}{2}}\dfrac{\cos^2x}{5-4\sin x}\text{d}x\right]\\
=&2\int_0^{\frac{\pi}{2}}\left[\dfrac{\sin^2x}{5+4\cos x}+\dfrac{\sin^2x}{5-4\cos x}\right]\text{d}x\\
=&2\int_0^{\frac{\pi}{2}}\dfrac{10\sin^2x}{25-16\cos^2x}\text{d}x\\
=&20\int_0^{\frac{\pi}{2}}\dfrac{1-\cos^2x}{25-16\cos^2x}\text{d}x\\
=&\dfrac{20}{16}\int_{0}^{\frac{\pi}{2}}\dfrac{25-16\cos^2x-9}{25-16\cos^2x}\text{d}x\\
=&\dfrac{5}{4}\int_{0}^{\frac{\pi}{2}}\left[1-\dfrac{9}{25-16\cos^2x}\right]\text{d}x\\
=&\dfrac{5\pi}{8}-\dfrac{45}{4}\int_{0}^{\frac{\pi}{2}}\dfrac{1}{25\sec^2x-16}\text{d}(\tan x)\\
=&\dfrac{5\pi}{8}-\dfrac{45}{4}\int_0^{\infty}\dfrac{1}{25x^2+9}\text{d}x\\
=&\dfrac{5\pi}{8}-\dfrac{45}{4}\dfrac{\pi}{30}\\
=&\dfrac{\pi}{4} \end{aligned}$

这题也可以用留数法