Page:Gauss - Théorie du mouvement des corps célestes, traduction Dubois, 1864.djvu/348

329

NOTE IV.

On a

	$\log 54,80779$	${}={}$	1,7388423	
	$\mathrm {c^{t}} \log 6,590997$	${}={}$	1,1810489
	$\mathrm {c^{t}} {\tfrac {3}{2}}\log 10$	${}={}$	2,5000000
	$\log \operatorname {tang} \mathrm {B}$	${}={}$	1,4198912
	 $\mathrm {B}$	${}={}$	114° 43′ 58,82″
	${\tfrac {1}{2}}\mathrm {B}$	${}={}$	107° 21′ 59,41″
	$\log \operatorname {tang} {\tfrac {1}{2}}\mathrm {B}$	${}={}$	1,1115410
	$\log \operatorname {tang} \mathrm {A}$	${}={}$	1,7038470
	 $\mathrm {A}$	${}={}$	126° 49′ 23,82″
	 $2\mathrm {A}$	${}={}$	153° 38′ 47,64″
	$\log \operatorname {cotang} 2\mathrm {A}$	${}={}$	1,8668836
	$\log 2$	${}={}$	0,301030
	$\log \operatorname {tang} {\tfrac {1}{2}}v$	${}={}$	0,1679136
	${\tfrac {1}{2}}v$	${}={}$	155° 48′ 36,755″
d’où	$v$	${}={}$	111° 37′ 13,51″

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Note IV (art. 94).

En posant $\operatorname {tang} {\frac {1}{2}}v=\theta ,$ on a

$dv={\frac {2\,d\theta }{1+\theta ^{2}}},$ et  $\cos v={\frac {1-\theta ^{2}}{1+\theta ^{2}}}\,;$

d’où

$\int {\frac {(1+e)^{\frac {3}{2}}}{(1+e\cos v)^{2}{\sqrt {2}}}}$	${}=\int {\frac {(1+e)^{\frac {3}{2}}2\,d\theta (1+\theta ^{2})}{\left(1+\theta ^{2}\right)^{2}{\sqrt {2}}\left(1+{\dfrac {e(1-\theta ^{2})}{1+\theta ^{2}}}\right)^{2}}},$
	${}=\int {\frac {(1+e)^{\frac {3}{2}}2\,d\theta (1+\theta ^{2})}{\left(1+e\right)^{2}{\sqrt {2}}\left(1+{\dfrac {(1-e)}{(1+e)}}\theta ^{2}\right)^{2}}},$
	${}=\int {\frac {(1+e)^{-{\frac {1}{2}}}{\sqrt {2}}\,d\theta (1+\theta ^{2})}{\left(1+{\dfrac {(1-e)}{(1+e)}}\theta ^{2}\right)^{2}}},$