積分　1/sinx

$3$\displaystyle{ \int \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sin x \hspace{2}}dx }$

$3$\displaystyle{ \int \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sin x \hspace{2}}dx }$ $3$\displaystyle{ =\int \frac{\hspace{2} \sin x \hspace{2}}{\hspace{2} { \sin }^{2}x \hspace{2}}dx }$ $3$\displaystyle{ =\int \frac{\hspace{2} \sin x \hspace{2}}{\hspace{2} 1- { \cos }^{2}x \hspace{2}}dx }$

$3$\displaystyle{f $ \cos x $ \sin x}$ の形に式が変形できたので， $3$\displaystyle{\cos x=t}$ とおいて置換積分を行う．

$3$\displaystyle{\frac{\hspace{2} dt \hspace{2}}{\hspace{2} dx \hspace{2}}=- \sin x\rightarrow \sin xdx=- dt}$

となるので

$3$\displaystyle{ \int \frac{\hspace{2} \sin x \hspace{2}}{\hspace{2} 1- { \cos }^{2}x \hspace{2}}dx }$ $3$\displaystyle{ =\int \frac{\hspace{2} - dt \hspace{2}}{\hspace{2} 1- {t}^{2} \hspace{2}} }$ $3$\displaystyle{ =\int \frac{\hspace{2} - dt \hspace{2}}{\hspace{2} \left({ 1- t }\right)\left({ 1+t }\right) \hspace{2}} }$

分数関数の積分になるので，部分分数に分解をする．

$3$\displaystyle{ \frac{\hspace{2} - 1 \hspace{2}}{\hspace{2} \left({ 1- t }\right)\left({ 1+t }\right) \hspace{2}} }$ $3$\displaystyle{ =\frac{\hspace{2}A\hspace{2}}{\hspace{2} 1- t \hspace{2}}+\frac{\hspace{2}B\hspace{2}}{\hspace{2} 1+t \hspace{2}} }$ $3$\displaystyle{ =\frac{\hspace{2} A\left({ 1+t }\right)+B\left({ 1- t }\right) \hspace{2}}{\hspace{2} \left({ 1- t }\right)\left({ 1+t }\right) \hspace{2}} }$ $3$\displaystyle{ =\frac{\hspace{2} \left({ A- B }\right)t+\left({ A+B }\right) \hspace{2}}{\hspace{2} \left({ 1- t }\right)\left({ 1+t }\right) \hspace{2}} }$

$3$\displaystyle{ \left{{ \begin{array}{lll} A- B=0 & \vspace{6}\\ A+B=- 1 & \vspace{6}\\ \end{array} }$

$3$\displaystyle{ A=B }$

$3$\displaystyle{ 2B=- 1 }$

$3$\displaystyle{ B=- \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}},A=- \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} }$

$3$\displaystyle{ \frac{\hspace{2} - 1 \hspace{2}}{\hspace{2} \left({ 1- t }\right)\left({ 1+t }\right) \hspace{2}} }$ $3$\displaystyle{ =- \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\left({ \frac{\hspace{2}1\hspace{2}}{\hspace{2} 1- t \hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2} 1+t \hspace{2}} }\right) }$

よって

$3$\displaystyle{ \int \frac{\hspace{2} - dt \hspace{2}}{\hspace{2} \left({ 1- t }\right)\left({ 1+t }\right) \hspace{2}} }$ $3$\displaystyle{ =- \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\int \left({ \frac{\hspace{2}1\hspace{2}}{\hspace{2} 1- t \hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2} 1+t \hspace{2}} }\right)dt }$

$3$\displaystyle{ =- \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\left{{ - \log \left|{ 1- t }\right|+\log \left|{ 1+t }\right| }\right}+C }$

$3$\displaystyle{ =- \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\log \left|{ \frac{\hspace{2} 1+t \hspace{2}}{\hspace{2} 1- t \hspace{2}} }\right|+C }$

$3$\displaystyle{ =\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\log \left|{ \frac{\hspace{2} 1- t \hspace{2}}{\hspace{2} 1+t \hspace{2}} }\right|+C }$

$3$\displaystyle{ =\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\log \left({ \frac{\hspace{2} 1- \cos x \hspace{2}}{\hspace{2} 1+\cos x \hspace{2}} }\right)+C }$ 　　（ $3$\displaystyle{C}$ は積分定数）

半角の公式をもちいると

$3$\displaystyle{ \int \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sin x \hspace{2}}dx }$ $3$\displaystyle{ =\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\log \frac{\hspace{2} { \sin }^{2}\frac{\hspace{2}x\hspace{2}}{\hspace{2}2\hspace{2}} \hspace{2}}{\hspace{2} { \cos }^{2}\frac{\hspace{2}x\hspace{2}}{\hspace{2}2\hspace{2}} \hspace{2}}+C }$

$3$\displaystyle{ =\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\log \left({ { \tan }^{2}\frac{\hspace{2}x\hspace{2}}{\hspace{2}2\hspace{2}} }\right)+C }$

$3$\displaystyle{ =\log \left|{ \tan \frac{\hspace{2}x\hspace{2}}{\hspace{2}2\hspace{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

積分 1/sinx

積分　1/sinx