更新积分

advanced-math/exercise/4-integal-of-functions-of-single-variable/integal-of-functions-of-single-variable.tex

@@ -391,7 +391,7 @@ $\int(x^3+2x+6)e^{2x}\,\textrm{d}x=\int x^3e^{2x}\,\textrm{d}x+2\int xe^{2x}\,\t
     $x^3+2x+6$ & $3x^2+2$ & $6x$ & 6 & 0 \\ \hline
     $e^{2x}$ & $\dfrac{1}{2}e^{2x}$ & $\dfrac{1}{4}e^{2x}$ & $\dfrac{1}{8}e^{2x}$ & $\dfrac{1}{16}e^{2x}$ \\
     \hline
-\end{tabular}
+\end{tabular}\medskip
 
 $\therefore=(x^3+2x+6)\left(\dfrac{1}{2}e^{2x}\right)-(3x^2+2)\left(\dfrac{1}{4}e^{2x}\right)+6x\left(\dfrac{1}{8}e^{2x}\right)-6\left(\dfrac{1}{16}e^{2x}\right)+\displaystyle{\int0\cdot(\dfrac{1}{16}e^{2x})\,\textrm{d}x}=\left(\dfrac{1}{2}x^3-\dfrac{3}{4}x^2+\dfrac{7}{4}x+\dfrac{17}{8}\right)e^{2x}+C$。
 

advanced-math/knowledge/2-derivatives-and-differential/derivatives-and-differential.tex

@@ -581,7 +581,7 @@ $
         $\sin x$ & $\cos x$ & $\cos x$ & $-\sin x$ \\ \hline
         $\tan x$ & $\dfrac{1}{\cos^2x}=\sec^2x$ & $\cot x$ & $\dfrac{1}{\sin^2x}=\csc^2x$ \\ \hline
         $\sec x$ & $\sec x\tan x$ & $\csc x$ & $-\csc x\cot x$ \\ \hline
-        $\arcsin x$ & $\dfrac{1}{1-x^2}$ & $\arccos x$ & $-\dfrac{1}{1-x^2}$ \\ \hline
+        $\arcsin x$ & $\dfrac{1}{\sqrt{1-x^2}}$ & $\arccos x$ & $-\dfrac{1}{\sqrt{1-x^2}}$ \\ \hline
         $\arctan x$ & $\dfrac{1}{1+x^2}$ & $\textrm{arccot}\,x$ & $-\dfrac{1}{1+x^2}$ \\ \hline
         $\textrm{arcsec}\,x$ & $\dfrac{1}{x\sqrt{x^2-1}}$ & $\textrm{arccsc}\,x$ & $-\dfrac{1}{x\sqrt{x^2-1}}$ \\
         \hline

advanced-math/knowledge/4-indefinite-integral-and-definite-integral/indefinite-integral-and-definite-integral.tex

@@ -597,6 +597,8 @@ $\int_0^\pi xf(\sin x)\,\textrm{d}x=-\int_\pi^0(\pi-t)f(\sin(\pi-t))\,\textrm{d}
 
 当积分区间为无穷区间，或被积函数为无界函数，那么定积分就无法“定”下来，所以这种积分就是反常积分。
 
+对于无穷区间的反常积分首先求出原函数，然后代入上下限。
+
 \subsubsection{无穷区间}
 
 设函数$f(x)$在区间$[a,+\infty)$上连续，任取$t>a$，做定积分$\int_a^tf(x)\,\textrm{d}x$，对这种变上限积分的极限$\lim\limits_{t\to+\infty}\int_a^tf(x)\,\textrm{d}x$就是$f(x)$在无穷区间$[a,+\infty)$上的反常积分，记为$\int_a^{+\infty}f(x)\,\textrm{d}x$。
@@ -623,7 +625,7 @@ $\int_0^\pi xf(\sin x)\,\textrm{d}x=-\int_\pi^0(\pi-t)f(\sin(\pi-t))\,\textrm{d}
 
 对于无界函数的反常积分要求的就是$\lim\limits_{x\to a}f(x)$。
 
-\subsection{* 反常积分的判敛}
+% \subsection{* 反常积分的判敛}
 
 \subsection{不定积分与定积分的区别与联系}