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# tan(x/2)=t とおく置換積分

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

より

• $3\displaystyle{ \int f\(\sin x,\text{\hspace{5}}\cos x\)dx }$

• $3\displaystyle{ = \int f\( \frac{\hspace{2} 2t \hspace{2}}{\hspace{2} 1+{t}^{2} \hspace{2}},\frac{\hspace{2} 1- {t}^{2} \hspace{2}}{\hspace{2} 1+{t}^{2} \hspace{2}} \)\cdot\hspace{5}\frac{\hspace{2}2\hspace{2}}{\hspace{2} 1+{t}^{2} \hspace{2}}dt }$

となる．

## ■$3\displaystyle{\cos x=\frac{\hspace{2} 1- {t}^{2} \hspace{2}}{\hspace{2} 1+{t}^{2} \hspace{2}}}$  を導く

2倍角の公式と，$3\displaystyle{\tan \frac{\hspace{2}x\hspace{2}}{\hspace{2}2\hspace{2}}=t}$  より，

$3\displaystyle{\tan x=\frac{\hspace{2} 2\tan \frac{\hspace{2}x\hspace{2}}{\hspace{2}2\hspace{2}} \hspace{2}}{\hspace{2} 1- { \tan }^{2}\frac{\hspace{2}x\hspace{2}}{\hspace{2}2\hspace{2}} \hspace{2}}=\frac{\hspace{2} 2t \hspace{2}}{\hspace{2} 1- {t}^{2} \hspace{2}}}$

$3\displaystyle{{\tan }^{2}\frac{\hspace{2}x\hspace{2}}{\hspace{2}2\hspace{2}}+1=\frac{\hspace{2}1\hspace{2}}{\hspace{2} { \cos }^{2}\frac{\hspace{2}x\hspace{2}}{\hspace{2}2\hspace{2}} \hspace{2}}}$

よって，

$3\displaystyle{{\cos }^{2}\frac{\hspace{2}x\hspace{2}}{\hspace{2}2\hspace{2}}=\frac{\hspace{2}1\hspace{2}}{\hspace{2} { \tan }^{2}\frac{\hspace{2}x\hspace{2}}{\hspace{2}2\hspace{2}}+1 \hspace{2}}=\frac{\hspace{2}1\hspace{2}}{\hspace{2} 1+{t}^{2} \hspace{2}}}$

$3\displaystyle{{\cos }^{2}\frac{\hspace{2}x\hspace{2}}{\hspace{2}2\hspace{2}}=\frac{\hspace{2} 1+\cos x \hspace{2}}{\hspace{2}2\hspace{2}}}$

したがって

$3\displaystyle{ \cos x }$

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

$3\displaystyle{ =\frac{\hspace{2}2\hspace{2}}{\hspace{2} 1+ {t}^{2} \hspace{2}}- 1 }$

$3\displaystyle{ =\frac{\hspace{2} 1- {t}^{2} \hspace{2}}{\hspace{2} 1+ {t}^{2} \hspace{2}} }$

## ■$3\displaystyle{\sin x=\frac{\hspace{2} 2t \hspace{2}}{\hspace{2} 1+{t}^{2} \hspace{2}}}$  を導く

$3\displaystyle{ \sin x }$

$3\displaystyle{ =\tan x\cos x }$

$3\displaystyle{ =\frac{\hspace{2} 2t \hspace{2}}{\hspace{2} 1- {t}^{2} \hspace{2}}\cdot\hspace{5}\frac{\hspace{2} 1- {t}^{2} \hspace{2}}{\hspace{2} 1+ {t}^{2} \hspace{2}} }$

$3\displaystyle{ =\frac{\hspace{2} 2t \hspace{2}}{\hspace{2} 1+ {t}^{2} \hspace{2}} }$

## ■$3\displaystyle{dx=\frac{\hspace{2}2\hspace{2}}{\hspace{2} 1+{t}^{2} \hspace{2}}dt}$ を導く

$3\displaystyle{t=\tan \frac{\hspace{2}x\hspace{2}}{\hspace{2}2\hspace{2}}=\frac{\hspace{2} \sin \frac{\hspace{2}x\hspace{2}}{\hspace{2}2\hspace{2}} \hspace{2}}{\hspace{2} \cos \frac{\hspace{2}x\hspace{2}}{\hspace{2}2\hspace{2}} \hspace{2}}}$  なので，

$3\displaystyle{ \frac{\hspace{2} dt \hspace{2}}{\hspace{2} dx \hspace{2}} }$

$3\displaystyle{ =\frac{\hspace{2} { \( \sin \frac{\hspace{2}x\hspace{2}}{\hspace{2}2\hspace{2}} \) }^{\prime }\cos \frac{\hspace{2}x\hspace{2}}{\hspace{2}2\hspace{2}}- \sin \frac{\hspace{2}x\hspace{2}}{\hspace{2}2\hspace{2}}{ \( \cos \frac{\hspace{2}x\hspace{2}}{\hspace{2}2\hspace{2}} \) }^{\prime } \hspace{2}}{\hspace{2}{ \( \cos \frac{\hspace{2}x\hspace{2}}{\hspace{2}2\hspace{2}} \) }^{2}\hspace{2}} }$

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

$3\displaystyle{ =\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\cdot\hspace{5}\frac{\hspace{2}1\hspace{2}}{\hspace{2} {\cos }^{2}\frac{\hspace{2}x\hspace{2}}{\hspace{2}2\hspace{2}} \hspace{2}} }$

$3\displaystyle{ =\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\cdot\hspace{5}\frac{\hspace{2}1\hspace{2}}{\hspace{2}\frac{\hspace{2}1\hspace{2}}{\hspace{2} 1+ {t}^{2} \hspace{2}}\hspace{2}} }$

$3\displaystyle{ =\frac{\hspace{2} 1+ {t}^{2} \hspace{2}}{\hspace{2}2\hspace{2}} }$

したがって，

$3\displaystyle{dx=\frac{\hspace{2}2\hspace{2}}{\hspace{2} 1+{t}^{2} \hspace{2}}dt}$

 ・$3\displaystyle{\int \frac{\hspace{2}1\hspace{2}}{\hspace{2} \cos x \hspace{2}}dx }$ ⇒解説と解答 ・$3\displaystyle{\int \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sin x \hspace{2}}dx }$ ⇒解説と解答 ・$3\displaystyle{\int \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sin x+\cos x+1 \hspace{2}}dx }$ ⇒解説と解答

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