积分练习 1
积分练习 1
1
求积分 $$ I=\int_0^{\frac{\pi}{2}}\sin x\ln\sin x\text{d}x $$
【Sol】:
$\begin{aligned}
I=&\int_0^{\frac{\pi}{2}}\cos x\ln\cos x\text{d}x\\
=&\int_0^{\frac{\pi}{2}}\ln\cos x\text{d}(\sin x-1)\\
=&(\sin x-1)\ln\cos x\bigg|_{0}^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}}(\sin x-1)\text{d}(\ln \cos x)\\
=&0-0-\int_0^{\frac{\pi}{2}}\sin x\text{d}(\ln\cos x)+\ln\cos x\bigg|_0^{\frac{\pi}{2}}
\end{aligned}$
令 $\displaystyle J=-\int_0^{\frac{\pi}{2}}\sin x\text{d}(\ln\cos x)$
$\begin{aligned}
J=&\int_0^{\frac{\pi}{2}}\sin x\tan x\text{d}x\\
=&\int_0^{\frac{\pi}{2}}\dfrac{\sin^2x}{\cos x}\text{d}x\\
=&\int_0^{\frac{\pi}{2}}\dfrac{1-\cos^2x}{\cos x}\text{d}x\\
=&\left[\ln|\sec x+\tan x|-\sin x\right]\bigg|_0^{\frac{\pi}{2}}
\end{aligned}$
因此
$\begin{aligned}
I=&J+\ln\cos x\bigg|_ 0^{\frac{\pi}{2}}\\
=&\left[\ln|\sec x+\tan x|-\sin x\right] \bigg|_{0}^{ \frac{\pi}{2}} +\ln\cos x\bigg|_{0}^{\frac{\pi}{2}}\\
=&\left[\ln|1+\sin x|-\sin x\right]\bigg|_0^{\frac{\pi}{2}}\\
=&\ln 2-1
\end{aligned}$
2
$$ I=\int_0^1\dfrac{x\ln x}{\sqrt{1-x^2}}\text{d}x $$
【Sol】: 令 $x=\sin t$ 即和上一题一样
或者
$\begin{aligned}
I=&\int_0^1\dfrac{\sqrt{x}\ln\sqrt{x}}{\sqrt{1-x}}\text{d}(\sqrt{x})\\
=&\dfrac{1}{4}\int_0^1\dfrac{\ln x}{\sqrt{1-x}}\text{d}x\\
=&\dfrac{1}{4}\int_{0}^{1}\dfrac{\ln(1-x)}{\sqrt{x}}\text{d}x\\
=&\dfrac{1}{4}\int_{0}^{1}\dfrac{\ln(1-x^2)}{x}\text{d}(x^2)\\
=&\dfrac{1}{2}\int_0^1\ln(1-x^2)\text{d}x\\
=&\dfrac{1}{2}\int_{0}^{1}\ln(1-x^2)\text{d}(x-1)\\
=&\dfrac{1}{2}\left[\ln(1-x^2)(x-1)\bigg|_{0}^{1}-\int_0^1(x-1)\dfrac{-2x}{1-x^2}\text{d}x \right]\\
=&\int_{0}^{1}\dfrac{-x}{1+x}\text{d}x=\int_{0}^{1}\left[\dfrac{1}{1+x}-1 \right]\text{d}x\\
=&\ln2-1
\end{aligned}$
方法二主要是变量代换和区间再现
3
求积分 $$ I=\int_0^1\sin\pi x\ln\Gamma(x)\text{d}x $$
【Sol】:
$\begin{aligned}
I=&\int_0^1\sin\pi(1-x)\ln\Gamma(1-x)\text{d}x\\
=&\dfrac{1}{2}\int_{0}^{1}\sin\pi x\ln(\Gamma(x)\Gamma(1-x))\text{d}x\\
=&\dfrac{1}{2}\int_{0}^{1}\sin\pi x\ln\left(\dfrac{\pi}{\sin\pi x}\right)\text{d}x\\
=&\dfrac{1}{2}\int_0^{\pi}\sin x\ln\left(\dfrac{\pi}{\sin x}\right)\text{d}\left(\dfrac{x}{\pi}\right)\\
=&\dfrac{1}{\pi}\int_{0}^{\frac{\pi}{2}}\sin x\ln\left(\dfrac{\pi}{\sin x}\right)\text{d}x\\
=&\dfrac{1}{\pi}\ln\pi\int_0^{\frac{\pi}{2}}\sin x\text{d}x-\dfrac{1}{\pi}\int_0^{\frac{\pi}{2}}\sin x\ln\sin x\text{d}x\\
=&\dfrac{1}{\pi}-\dfrac{1}{\pi}(\ln2-1)\\
=&\dfrac{1}{\pi}\left[\ln\dfrac{\pi}{2}+1 \right]
\end{aligned}$
这题用到了余元公式和区间再现
4
求积分 $$ I=\int_{0}^{\frac{\pi}{2}}\dfrac{\sin x\cos x}{\sin x+\cos x} $$
【Sol】:
$\begin{aligned}
I=&\int_{0}^{\frac{\pi}{2}}\dfrac{\sin x\cos x}{\sin x+\cos x}\text{d}x\\
=&\dfrac{1}{2}\int_{0}^{\frac{\pi}{2}}\dfrac{(\sin x+\cos x)^2-1}{\sin x+\cos x}\text{d}x\\
=&\dfrac{1}{2}\int_0^{\frac{\pi}{2}}\left[\sin x+\cos x-\dfrac{1}{\sin x+\cos x}\right]\text{d}x\\
=&\dfrac{1}{2}\left[(\sin x-\cos x)\bigg|_{0}^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}}\dfrac{1}{\sqrt{2}\cos\left(x-\dfrac{\pi}{4}\right)}\text{d}x\right]\\
=&\dfrac{1}{2}\left[2-\dfrac{1}{\sqrt{2}}\int_{0-\frac{\pi}{4}}^{\frac{\pi}{2}-\frac{\pi}{4}}\sec x\text{d}x\right]\\
=&1-\dfrac{1}{2\sqrt{2}}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sec x\text{d}x\\
=&1-\dfrac{1}{\sqrt{2}}\int_{0}^{\frac{\pi}{4}}\sec x\text{d}x\\
=&1-\dfrac{1}{\sqrt{2}}\ln|\sec x+\tan x|\bigg|_{0}^{\frac{\pi}{4}}\\
=&1-\dfrac{1}{\sqrt{2}}\ln|\sqrt{2}+1|=1+\dfrac{1}{\sqrt{2}}\ln|\sqrt{2}-1|
\end{aligned}$
来自 函数君爱复艾克斯