POINT
- 積分公式の一覧（途中計算あり）．
- 対象：三角関数・双曲線関数・指数関数・対数関数．
基本的な不定積分公式を導出します．
以下では$a>0$とし，積分定数は省略します．
他の積分公式はこちら：
指数関数
$(e^x)^\prime=e^x$なので\begin{aligned}
\int e^x\,\mathrm{d}x
&=e^x.
\end{aligned}
これを用いれば\int e^x\,\mathrm{d}x
&=e^x.
\end{aligned}
\begin{aligned}
\int a^x\,\mathrm{d}x
&=\int e^{\log a^x}\,\mathrm{d}x \\
&=\int e^{x\log a}\,\mathrm{d}x \\
&=\frac{a^x}{\log a}
\end{aligned}
となります．\int a^x\,\mathrm{d}x
&=\int e^{\log a^x}\,\mathrm{d}x \\
&=\int e^{x\log a}\,\mathrm{d}x \\
&=\frac{a^x}{\log a}
\end{aligned}
三角関数
以下では，関数 $\displaystyle\sec x=\frac{1}{\cos x}$, $\displaystyle\mathrm{cosec\,} x=\frac{1}{\sin x}$, $\displaystyle\cot x=\frac{1}{\tan x}$を使います．三角関数の導関数（$(\sin x)^\prime=\cos x$, $(\cos x)^\prime=-\sin x$）から
\begin{aligned}
\int \sin x\,\mathrm{d}x
&=-\cos x\\
\int \cos x\,\mathrm{d}x
&=\sin x\\
%
\int \tan x\,\mathrm{d}x
&=-\int \frac{(\cos x)^\prime}{\cos x}\,\mathrm{d}x \\
&=-\log|\cos x|
\end{aligned}
がわかります．\int \sin x\,\mathrm{d}x
&=-\cos x\\
\int \cos x\,\mathrm{d}x
&=\sin x\\
%
\int \tan x\,\mathrm{d}x
&=-\int \frac{(\cos x)^\prime}{\cos x}\,\mathrm{d}x \\
&=-\log|\cos x|
\end{aligned}
\begin{aligned}
&\int \sec x \,\mathrm{d}x = \int \frac{1}{\cos x} \,\mathrm{d}x \\
&=\log\Biggl| \frac{1+\sin x}{1- \sin x}\Biggr|^{1/2} \\
&=\log|\sec x+\tan x|
\end{aligned}
&\int \sec x \,\mathrm{d}x = \int \frac{1}{\cos x} \,\mathrm{d}x \\
&=\log\Biggl| \frac{1+\sin x}{1- \sin x}\Biggr|^{1/2} \\
&=\log|\sec x+\tan x|
\end{aligned}
\begin{aligned}
&\int \sec x \,\mathrm{d}x = \int \frac{1}{\cos x} \,\mathrm{d}x \\
&=\int \frac{\cos x}{1-\sin^2x} \,\mathrm{d}x
=\int \frac{\mathrm{d}(\sin x)}{1-\sin^2x} \\
&=-\int\Biggl(\frac{1}{y-1} -\frac{1}{y+1}\Biggr) \frac{\mathrm{d}y}{2}
=\frac{1}{2}\log\Biggl| \frac{y+1}{y-1}\Biggr| \\
&=\log\Biggl| \frac{1+\sin x}{1- \sin x}\Biggr|^{1/2} \\
&=\log\Biggl| \frac{(1+\sin x)^2}{(1-\sin x)(1+\sin x)}\Biggr|^{1/2}
=\log\Biggl| \frac{1+\sin x}{\cos x}\Biggr| \\
&=\log|\sec x+\tan x|
\end{aligned}
となる．//&\int \sec x \,\mathrm{d}x = \int \frac{1}{\cos x} \,\mathrm{d}x \\
&=\int \frac{\cos x}{1-\sin^2x} \,\mathrm{d}x
=\int \frac{\mathrm{d}(\sin x)}{1-\sin^2x} \\
&=-\int\Biggl(\frac{1}{y-1} -\frac{1}{y+1}\Biggr) \frac{\mathrm{d}y}{2}
=\frac{1}{2}\log\Biggl| \frac{y+1}{y-1}\Biggr| \\
&=\log\Biggl| \frac{1+\sin x}{1- \sin x}\Biggr|^{1/2} \\
&=\log\Biggl| \frac{(1+\sin x)^2}{(1-\sin x)(1+\sin x)}\Biggr|^{1/2}
=\log\Biggl| \frac{1+\sin x}{\cos x}\Biggr| \\
&=\log|\sec x+\tan x|
\end{aligned}
\begin{aligned}
&\int \mathrm{cosec\,} x\,\mathrm{d}x = \int \frac{1}{\sin x} \,\mathrm{d}x \\
&=\log\Biggl| \frac{1-\cos x}{1+\cos x}\Biggr|^{1/2} \\
&=\log|\mathrm{cosec\,} x- \cot x| \\
&=\log|\tan (x/2)|
\end{aligned}
&\int \mathrm{cosec\,} x\,\mathrm{d}x = \int \frac{1}{\sin x} \,\mathrm{d}x \\
&=\log\Biggl| \frac{1-\cos x}{1+\cos x}\Biggr|^{1/2} \\
&=\log|\mathrm{cosec\,} x- \cot x| \\
&=\log|\tan (x/2)|
\end{aligned}
\begin{aligned}
&( \tan x )^\prime = \frac{\mathrm{d}}{\mathrm{d}x} \biggl(\frac{\sin x}{\cos x} \biggr) \\
&=\cos x\frac{1}{\cos x} + \sin x \biggl(-\frac{1}{\cos^2 x}\biggr)(-\sin x) \\
&=\frac{1}{\cos^2 x}
\end{aligned}
より&( \tan x )^\prime = \frac{\mathrm{d}}{\mathrm{d}x} \biggl(\frac{\sin x}{\cos x} \biggr) \\
&=\cos x\frac{1}{\cos x} + \sin x \biggl(-\frac{1}{\cos^2 x}\biggr)(-\sin x) \\
&=\frac{1}{\cos^2 x}
\end{aligned}
\begin{aligned}
\frac{( \tan x )^\prime}{\tan x}
&=\frac{1}{\sin x \cos x} \\
&=\frac{2}{\sin (2x)}
\end{aligned}
であることを利用すれば，\frac{( \tan x )^\prime}{\tan x}
&=\frac{1}{\sin x \cos x} \\
&=\frac{2}{\sin (2x)}
\end{aligned}
\begin{aligned}
&\int \mathrm{cosec\,} x\,\mathrm{d}x = \int \frac{1}{\sin x} \,\mathrm{d}x \\
&=\int \frac{[ \tan (x/2) ]^\prime}{\tan (x/2)} \,\mathrm{d}\biggl(\frac{x}{2}\biggl) \\
&=\log\Bigl| \tan (x/2) \Bigr|
\end{aligned}
と計算できる．また，&\int \mathrm{cosec\,} x\,\mathrm{d}x = \int \frac{1}{\sin x} \,\mathrm{d}x \\
&=\int \frac{[ \tan (x/2) ]^\prime}{\tan (x/2)} \,\mathrm{d}\biggl(\frac{x}{2}\biggl) \\
&=\log\Bigl| \tan (x/2) \Bigr|
\end{aligned}
\begin{aligned}
&\int \mathrm{cosec\,} x\,\mathrm{d}x = \int \frac{1}{\sin x} \,\mathrm{d}x \\
&=\int \frac{\sin x}{1-\cos^2x} \,\mathrm{d}x
=\int \frac{\mathrm{d}(\cos x)}{\cos^2x-1} \\
&=\int\Biggl(\frac{1}{y-1} -\frac{1}{y+1}\Biggr) \frac{\mathrm{d}y}{2}
=\frac{1}{2}\log\Biggl| \frac{y-1}{y+1}\Biggr| \\
&=\log\Biggl| \frac{1-\cos x}{1+\cos x}\Biggr|^{1/2} \\
&=\log\Biggl| \frac{(1-\cos x)^2}{(1-\cos x)(1+\cos x)}\Biggr|^{1/2}
=\log\Biggl| \frac{1-\cos x}{\sin x}\Biggr| \\
&=\log|\mathrm{cosec\,} x- \cot x|
\end{aligned}
である．（半角の公式$\sin^2 \theta = (1-\cos 2\theta)/2$, $\cos^2 \theta = (1+\cos 2\theta)/2$を用いれば&\int \mathrm{cosec\,} x\,\mathrm{d}x = \int \frac{1}{\sin x} \,\mathrm{d}x \\
&=\int \frac{\sin x}{1-\cos^2x} \,\mathrm{d}x
=\int \frac{\mathrm{d}(\cos x)}{\cos^2x-1} \\
&=\int\Biggl(\frac{1}{y-1} -\frac{1}{y+1}\Biggr) \frac{\mathrm{d}y}{2}
=\frac{1}{2}\log\Biggl| \frac{y-1}{y+1}\Biggr| \\
&=\log\Biggl| \frac{1-\cos x}{1+\cos x}\Biggr|^{1/2} \\
&=\log\Biggl| \frac{(1-\cos x)^2}{(1-\cos x)(1+\cos x)}\Biggr|^{1/2}
=\log\Biggl| \frac{1-\cos x}{\sin x}\Biggr| \\
&=\log|\mathrm{cosec\,} x- \cot x|
\end{aligned}
\begin{aligned}
&\log\Biggl| \frac{1-\cos x}{1+\cos x}\Biggr|^{1/2} \\
&=\log\Biggl| \frac{2\sin^2(x/2)}{2\cos^2(x/2)}\Biggr|^{1/2} \\
&=\log|\tan (x/2)|
\end{aligned}
となる）．&\log\Biggl| \frac{1-\cos x}{1+\cos x}\Biggr|^{1/2} \\
&=\log\Biggl| \frac{2\sin^2(x/2)}{2\cos^2(x/2)}\Biggr|^{1/2} \\
&=\log|\tan (x/2)|
\end{aligned}
【参考】
\begin{aligned}
&( \cot x )^\prime = \frac{\mathrm{d}}{\mathrm{d}x} \biggl(\frac{\cos x}{\sin x} \biggr) \\
&=-\sin x\frac{1}{\sin x} + \cos x \biggl(-\frac{1}{\sin^2 x}\biggr)\cos x \\
&=-\frac{1}{\sin^2 x}
\end{aligned}
より&( \cot x )^\prime = \frac{\mathrm{d}}{\mathrm{d}x} \biggl(\frac{\cos x}{\sin x} \biggr) \\
&=-\sin x\frac{1}{\sin x} + \cos x \biggl(-\frac{1}{\sin^2 x}\biggr)\cos x \\
&=-\frac{1}{\sin^2 x}
\end{aligned}
\begin{aligned}
\frac{( \cot x )^\prime}{\cot x}
&=-\frac{1}{\sin x \cos x} \\
&=-\frac{2}{\sin (2x)}
\end{aligned}
であることを利用すれば，\frac{( \cot x )^\prime}{\cot x}
&=-\frac{1}{\sin x \cos x} \\
&=-\frac{2}{\sin (2x)}
\end{aligned}
\begin{aligned}
&\int \mathrm{cosec\,}x \,\mathrm{d}x = \int \frac{1}{\sin x} \,\mathrm{d}x \\
&=-\int \frac{[ \cot (x/2) ]^\prime}{\cot (x/2)} \,\mathrm{d}\biggl(\frac{x}{2}\biggl) \\
&=-\log\Bigl| \cot (x/2) \Bigr| \\
&=\log\Bigl| \tan (x/2) \Bigr|
\end{aligned}
となる．//&\int \mathrm{cosec\,}x \,\mathrm{d}x = \int \frac{1}{\sin x} \,\mathrm{d}x \\
&=-\int \frac{[ \cot (x/2) ]^\prime}{\cot (x/2)} \,\mathrm{d}\biggl(\frac{x}{2}\biggl) \\
&=-\log\Bigl| \cot (x/2) \Bigr| \\
&=\log\Bigl| \tan (x/2) \Bigr|
\end{aligned}
$\cot x$は$\tan x$と同じ方法で
\begin{aligned}
&\int \cot x\,\mathrm{d}x = \int \frac{1}{\tan x}\,\mathrm{d}x \\
&=\int \frac{(\sin x)^\prime}{\sin x}\,\mathrm{d}x \\
&=\log|\sin x|
\end{aligned}
と計算できる．&\int \cot x\,\mathrm{d}x = \int \frac{1}{\tan x}\,\mathrm{d}x \\
&=\int \frac{(\sin x)^\prime}{\sin x}\,\mathrm{d}x \\
&=\log|\sin x|
\end{aligned}
\begin{aligned}
&\int \sec^2x\,\mathrm{d}x = \int \frac{1}{\cos^2x} \,\mathrm{d}x \\
&=\int (\tan x)^\prime\,\mathrm{d}x \\
&=\tan x\\
&\int \mathrm{cosec\,}^2x\,\mathrm{d}x = \int \frac{1}{\sin^2x} \,\mathrm{d}x \\
&=-\int (\cot x)^\prime\,\mathrm{d}x \\
&=-\cot x
\end{aligned}
&\int \sec^2x\,\mathrm{d}x = \int \frac{1}{\cos^2x} \,\mathrm{d}x \\
&=\int (\tan x)^\prime\,\mathrm{d}x \\
&=\tan x\\
&\int \mathrm{cosec\,}^2x\,\mathrm{d}x = \int \frac{1}{\sin^2x} \,\mathrm{d}x \\
&=-\int (\cot x)^\prime\,\mathrm{d}x \\
&=-\cot x
\end{aligned}
双曲線関数
双曲線関数は，指数関数$e^x$を用いて\begin{aligned}
\sinh x
=\frac{e^x-e^{-x}}{2}
,\quad
\cosh x
=\frac{e^x + e^{-x}}{2}
\end{aligned}
で定義されます．簡単にわかるように
\sinh x
=\frac{e^x-e^{-x}}{2}
,\quad
\cosh x
=\frac{e^x + e^{-x}}{2}
\end{aligned}
- $\cosh^2 x -\sinh^2 x=1$
- $(\sinh x)^\prime = \cosh x$, $(\cosh x)^\prime = \sinh x$
が成り立ちます．
指数関数の積分公式（または上で触れた導関数）により，以下がわかります：
\begin{aligned}
&\int \sinh x\,\mathrm{d}x
=\cosh x \\
&\int \cosh x\,\mathrm{d}x
=\sinh x
\end{aligned}
&\int \sinh x\,\mathrm{d}x
=\cosh x \\
&\int \cosh x\,\mathrm{d}x
=\sinh x
\end{aligned}
対数関数
部分積分を行うことにより，対数関数の積分が計算できます：\begin{aligned}
\int \log x\,\mathrm{d}x
&=\int \Bigl[(x\log x)^\prime-x(\log x)^\prime\Bigr]\,\mathrm{d}x \\
&=x\log x-x
\end{aligned}
\int \log x\,\mathrm{d}x
&=\int \Bigl[(x\log x)^\prime-x(\log x)^\prime\Bigr]\,\mathrm{d}x \\
&=x\log x-x
\end{aligned}
有理関数
\begin{aligned}
\int x^r\,\mathrm{d}x
&=
\begin{cases}
\,\dfrac{x^{r+1}}{r+1}&r\neq-1\\
\,\log|x|& r=-1
\end{cases}
\end{aligned}
\int x^r\,\mathrm{d}x
&=
\begin{cases}
\,\dfrac{x^{r+1}}{r+1}&r\neq-1\\
\,\log|x|& r=-1
\end{cases}
\end{aligned}
$x=\tan y$と変数変換することにより
\begin{aligned}
\int\frac{\mathrm{d}x}{1+x^2}
&=\int\frac{\mathrm{d}(\tan y)}{1+(\tan y)^2} \\
&=\int\frac{1/\cos^2 y}{1/\cos^2 y}\,\mathrm{d}y \\
&=y \\
&=\arctan x.
\end{aligned}
この結果から，$1\rightarrow a$とした積分も\int\frac{\mathrm{d}x}{1+x^2}
&=\int\frac{\mathrm{d}(\tan y)}{1+(\tan y)^2} \\
&=\int\frac{1/\cos^2 y}{1/\cos^2 y}\,\mathrm{d}y \\
&=y \\
&=\arctan x.
\end{aligned}
\begin{aligned}
\int\frac{\mathrm{d}x}{x^2+a^2}
&=\int \frac{1}{a}\frac{\mathrm{d}(x/a)}{1+(x/a)^2} \\
&=\frac{1}{a}\arctan(x/a)
\end{aligned}
と計算できます．\int\frac{\mathrm{d}x}{x^2+a^2}
&=\int \frac{1}{a}\frac{\mathrm{d}(x/a)}{1+(x/a)^2} \\
&=\frac{1}{a}\arctan(x/a)
\end{aligned}
上で$a^2$の前の符号をマイナスに変えた積分は，部分分数分解が適用できる簡単な例になっています；
\begin{aligned}
\int\frac{\mathrm{d}x}{x^2-a^2}
&=\int\Bigl(\frac{-1}{x+a}+\frac{1}{x-a}\Bigr)\,\frac{\mathrm{d}x}{2a} \\
&=\frac{1}{2a}\log\Bigl|\frac{x-a}{x+a}\Bigr|
\end{aligned}
\int\frac{\mathrm{d}x}{x^2-a^2}
&=\int\Bigl(\frac{-1}{x+a}+\frac{1}{x-a}\Bigr)\,\frac{\mathrm{d}x}{2a} \\
&=\frac{1}{2a}\log\Bigl|\frac{x-a}{x+a}\Bigr|
\end{aligned}
無理関数
適当な変数変換により\begin{aligned}
\int\frac{\mathrm{d}x}{\sqrt{1-x^2}}
&=\int\frac{\mathrm{d}(\sin y)}{\sqrt{1-\sin^2y}} \\
&=\int\frac{\cos y}{|\cos y|}\,\mathrm{d}y \\
&=\arcsin x\\
\int\frac{\mathrm{d}x}{\sqrt{x^2+1}}
&=\int\frac{\mathrm{d}(\sinh y)}{\sqrt{(\sinh y)^2+1}} \\
&=\int\frac{\cosh y}{\cosh y}\,\mathrm{d}y \\
&=y \\
&=\log\bigl(x+\sqrt{x^2+1}\bigr)\\
\int\frac{\mathrm{d}x}{\sqrt{x^2-1}}
&=\int\frac{\mathrm{d}(\cosh y)}{\sqrt{(\cosh y)^2-1}} \\
&=\int\frac{\sinh y}{|\sinh y|}\,\mathrm{d}y \\
&=|y| \\
&=\log\bigl(x+\sqrt{x^2-1}\bigr),\quad(x>1)
\end{aligned}
と計算することができます．これらの結果を利用すれば\int\frac{\mathrm{d}x}{\sqrt{1-x^2}}
&=\int\frac{\mathrm{d}(\sin y)}{\sqrt{1-\sin^2y}} \\
&=\int\frac{\cos y}{|\cos y|}\,\mathrm{d}y \\
&=\arcsin x\\
\int\frac{\mathrm{d}x}{\sqrt{x^2+1}}
&=\int\frac{\mathrm{d}(\sinh y)}{\sqrt{(\sinh y)^2+1}} \\
&=\int\frac{\cosh y}{\cosh y}\,\mathrm{d}y \\
&=y \\
&=\log\bigl(x+\sqrt{x^2+1}\bigr)\\
\int\frac{\mathrm{d}x}{\sqrt{x^2-1}}
&=\int\frac{\mathrm{d}(\cosh y)}{\sqrt{(\cosh y)^2-1}} \\
&=\int\frac{\sinh y}{|\sinh y|}\,\mathrm{d}y \\
&=|y| \\
&=\log\bigl(x+\sqrt{x^2-1}\bigr),\quad(x>1)
\end{aligned}
\begin{aligned}
\int\frac{\mathrm{d}x}{\sqrt{a^2-x^2}}
&=\int\frac{\mathrm{d}(x/a)}{\sqrt{1-(x/a)^2}} \\
&=\arcsin (x/a) \\
\int\frac{\mathrm{d}x}{\sqrt{x^2+ a^2}}
&=\int\frac{\mathrm{d}(x/a)}{\sqrt{(x/a)^2+ 1}} \\
&=\log\biggl[(x/a)+\sqrt{(x/a)^2+1}\biggr]\\
\int\frac{\mathrm{d}x}{\sqrt{x^2- a^2}}
&=\int\frac{\mathrm{d}(x/a)}{\sqrt{(x/a)^2- 1}} \\
&=\log\biggl[(x/a)+\sqrt{(x/a)^2-1}\biggr] \\
&\qquad(x>a)
\end{aligned}
がわかります．さらにこれらの結果を用いれば，部分積分により\int\frac{\mathrm{d}x}{\sqrt{a^2-x^2}}
&=\int\frac{\mathrm{d}(x/a)}{\sqrt{1-(x/a)^2}} \\
&=\arcsin (x/a) \\
\int\frac{\mathrm{d}x}{\sqrt{x^2+ a^2}}
&=\int\frac{\mathrm{d}(x/a)}{\sqrt{(x/a)^2+ 1}} \\
&=\log\biggl[(x/a)+\sqrt{(x/a)^2+1}\biggr]\\
\int\frac{\mathrm{d}x}{\sqrt{x^2- a^2}}
&=\int\frac{\mathrm{d}(x/a)}{\sqrt{(x/a)^2- 1}} \\
&=\log\biggl[(x/a)+\sqrt{(x/a)^2-1}\biggr] \\
&\qquad(x>a)
\end{aligned}
\begin{aligned}
&\int\sqrt{a^2-x^2}\,\mathrm{d}x\\
&=\cdots\\
&=\frac{1}{2}\Bigl[ x\sqrt{a^2-x^2}+a^2\arcsin (x/a) \Bigr] \\
&\int\sqrt{x^2+ a^2}\,\mathrm{d}x\\
&=\cdots\\
&=\frac{1}{2}\Bigl[ x\sqrt{x^2+a^2}+a^2\log \Bigl|x+\sqrt{x^2+a^2}\Bigr| \Bigr] \\
&\int\sqrt{x^2- a^2}\,\mathrm{d}x\\
&=\cdots\\
&=\frac{1}{2}\Bigl[ x\sqrt{x^2-a^2}+a^2\log \Bigl|x+\sqrt{x^2-a^2}\Bigr| \Bigr]
\end{aligned}
が計算できます．&\int\sqrt{a^2-x^2}\,\mathrm{d}x\\
&=\cdots\\
&=\frac{1}{2}\Bigl[ x\sqrt{a^2-x^2}+a^2\arcsin (x/a) \Bigr] \\
&\int\sqrt{x^2+ a^2}\,\mathrm{d}x\\
&=\cdots\\
&=\frac{1}{2}\Bigl[ x\sqrt{x^2+a^2}+a^2\log \Bigl|x+\sqrt{x^2+a^2}\Bigr| \Bigr] \\
&\int\sqrt{x^2- a^2}\,\mathrm{d}x\\
&=\cdots\\
&=\frac{1}{2}\Bigl[ x\sqrt{x^2-a^2}+a^2\log \Bigl|x+\sqrt{x^2-a^2}\Bigr| \Bigr]
\end{aligned}
