積分　1/sinx

${\displaystyle \int \frac{1}{sinx}dx}$

${\displaystyle \int \frac{1}{\sin x}dx}$${\displaystyle =\int \frac{\sin x}{{\sin }^{2}x}dx}$${\displaystyle =\int \frac{\sin x}{1-{\cos }^{2}x}dx}$

${\displaystyle f\left(cosx\right)sinx}$  の形に式が変形できたので，${\displaystyle cosx=t}$  とおいて置換積分を行う．

${\displaystyle \frac{dt}{dx}=-sinx\to sinxdx=-dt}$

となるので

${\displaystyle \int \frac{\sin x}{1-{\cos }^{2}x}dx}$${\displaystyle =\int \frac{-dt}{1-t^2}}$${\displaystyle =\int \frac{-dt}{\left(1-t\right)\left(1+t\right)}}$

分数関数の積分になるので，部分分数に分解をする．

${\displaystyle \frac{-1}{\left(1-t\right)\left(1+t\right)}}$${\displaystyle =\frac{A}{1-t}+\frac{B}{1+t}}$${\displaystyle =\frac{A\left(1+t\right)+B\left(1-t\right)}{\left(1-t\right)\left(1+t\right)}}$${\displaystyle =\frac{\left(A-B\right)t+\left(A+B\right)}{\left(1-t\right)\left(1+t\right)}}$

${\displaystyle \left\{\begin{array}{l}A-B=0\hfill \\ A+B=-1\hfill \end{array}\right.}$

${\displaystyle A=B}$

${\displaystyle 2B=-1}$

${\displaystyle B=-\frac{1}{2},A=-\frac{1}{2}}$

${\displaystyle \frac{-1}{\left(1-t\right)\left(1+t\right)}}$${\displaystyle =-\frac{1}{2}\left(\frac{1}{1-t}+\frac{1}{1+t}\right)}$

よって

${\displaystyle \int \frac{-dt}{\left(1-t\right)\left(1+t\right)}}$${\displaystyle =-\frac{1}{2}\int \left(\frac{1}{1-t}+\frac{1}{1+t}\right)dt}$

${\displaystyle =-\frac{1}{2}\left\{-\log \left|1-t\right|+\log \left|1+t\right|\right\}+C}$

${\displaystyle =-\frac{1}{2}\log \left|\frac{1+t}{1-t}\right|+C}$

${\displaystyle =\frac{1}{2}\log \left|\frac{1-t}{1+t}\right|+C}$

${\displaystyle =\frac{1}{2}\log \left(\frac{1-\cos x}{1+\cos x}\right)+C}$　　（ ${\displaystyle C}$ は積分定数）

${\displaystyle \int \frac{1}{sinx}dx}$${\displaystyle =\frac{1}{2}log\left(\frac{1-cosx}{1+cosx}\right)+C}$

半角の公式をもちいると

${\displaystyle \int \frac{1}{\sin x}dx}$${\displaystyle =\frac{1}{2}\log \frac{{\sin }^2\frac{x}{2}}{{\cos }^2\frac{x}{2}}+C}$

${\displaystyle =\frac{1}{2}\log \left({\tan }^2\frac{x}{2}\right)+C}$

${\displaystyle =\log \left|\tan \frac{x}{2}\right|+C}$

また，別の置換方法を用いても解を得ることができる．
詳しくはここを参照．

 

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最終更新日：2023年1月30日