395
RÉSOLUES.
![{\displaystyle \operatorname {f} (a)=\int {\frac {\operatorname {d} a}{\sqrt {1-a^{2}}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/39a0da02648c48f5297cfd5173a969c01a8d306e)
changeant donc, tour à tour,
en
et
on aura, au moyen de notre théorème général, pour la somme de la suite proposée,
![{\displaystyle {\frac {1}{2}}\left\{\int {\frac {\operatorname {d} x\left(-\operatorname {Sin} .x+{\sqrt {-1}}\operatorname {Cos} .x\right)}{\sqrt {1-\operatorname {Cos} .2x-{\sqrt {-1}}\operatorname {Sin} .2x}}}-\int {\frac {\operatorname {d} x\left(\operatorname {Sin} .x+{\sqrt {-1}}\operatorname {Cos} .x\right)}{\sqrt {1-\operatorname {Cos} .2x+{\sqrt {-1}}\operatorname {Sin} .2x}}}\right\},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a6c893dad46a8693309f8dc3ddc720ea76a3746d)
ou encore
![{\displaystyle {\frac {1}{2}}\int \left\{-{\frac {\operatorname {d} x}{\sqrt {2\operatorname {Sin} .x}}}\left({\sqrt {\operatorname {Sin} .x-{\sqrt {-1}}\operatorname {Cos} .x}}+{\sqrt {\operatorname {Sin} .x+{\sqrt {-1}}\operatorname {Cos} .x}}\right)\right\},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4cd9d935cf047c94c13052cf4c56c73aa9b45a1)
ou bien
![{\displaystyle \int -{\frac {{\frac {1}{2}}\operatorname {d} x}{\sqrt {\operatorname {Sin} .x}}}.{\sqrt {1+\operatorname {Sin} .x}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/95fc462587e03235492b84b5e9885210cde98895)
Or, soit
, on aura
![{\displaystyle {\frac {{\frac {1}{2}}\operatorname {d} x\operatorname {Cos} .x}{\sqrt {\operatorname {Sin} .x}}}=\operatorname {d} t,\qquad {\frac {{\frac {1}{2}}\operatorname {d} x}{\sqrt {\operatorname {Sin} .x}}}={\frac {\operatorname {d} t}{\sqrt {1-t^{4}}}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8c144adb4ca4370c08f0159f59e3480bd14e42a3)
donc
![{\displaystyle \int -{\frac {{\frac {1}{2}}\operatorname {d} x}{\sqrt {\operatorname {Sin} .x}}}.{\sqrt {1+\operatorname {Sin} .x}}=\int -{\frac {\operatorname {d} t}{\sqrt {1-t^{4}}}}{\sqrt {1+t^{2}}}=\int -{\frac {\operatorname {d} t}{\sqrt {1-t^{2}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4d76f74d6a465c09ba90b6e5e3bd339b9add858)
![{\displaystyle =\operatorname {Arc} .(\operatorname {Cos} .=t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ba196720d16787d29f9258ee10d2bb9ef979bf12)
donc enfin la somme de la série proposée sera
![{\displaystyle \operatorname {Arc} .\left(\operatorname {Cos} .={\sqrt {\operatorname {Sin} .x}}\right)={\frac {1}{2}}\operatorname {Arc} .(\operatorname {Sin} .=2\operatorname {Sin} .x\operatorname {Cos} .x-1)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b0bdc07c5587b0089067d9526a18481ae926ef12)
comme nous l’avions également trouvé.