![{\displaystyle \int ^{m}a^{x}Cos.x\operatorname {d} x^{m}={\frac {a^{x}}{Log.^{m}a}}\times }](https://wikimedia.org/api/rest_v1/media/math/render/svg/dfbfac947a3ba8c7657339e40a249d87a872a06d)
![{\displaystyle \left(\operatorname {Cos} .x+{\frac {m}{1}}{\frac {\operatorname {Sin} .x}{\operatorname {Log} .a}}-{\frac {m}{1}}.{\frac {m+1}{2}}{\frac {\operatorname {Cos} .x}{\operatorname {Log} .^{2}a}}-{\frac {m}{1}}.{\frac {m+1}{2}}{\frac {m+2}{3}}{\frac {\operatorname {Sin} .x}{\operatorname {Log} .^{3}a}}+\ldots \right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/17976a0cf320df3d38d7d515439a46d753d9e5bb)
en faisant ensuite
on conclura de là
![{\displaystyle \int a^{x}\operatorname {Sin} .x\operatorname {d} x=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f2891837f0403ed06bd07436d6ef303a6ae0beba)
![{\displaystyle a^{x}\left({\frac {\operatorname {Sin} .x}{\operatorname {Log} .a}}-{\frac {\operatorname {Cos} .x}{\operatorname {Log} .^{2}a}}-{\frac {\operatorname {Sin} .x}{\operatorname {Log} .^{3}a}}+{\frac {\operatorname {Cos} .x}{\operatorname {Log} .^{4}a}}+{\frac {\operatorname {Sin} .x}{\operatorname {Log} .^{5}a}}-\ldots \right),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cdeca5477735eef7b2750b12181786df81ee9681)
![{\displaystyle \int a^{x}\operatorname {Cos} .x\operatorname {d} x=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3a7a7634d156634f2c922cd09c4b19933da52bcb)
![{\displaystyle a^{x}\left({\frac {\operatorname {Cos} .x}{\operatorname {Log} .a}}+{\frac {\operatorname {Sin} .x}{\operatorname {Log} .^{2}a}}-{\frac {\operatorname {Cos} .x}{\operatorname {Log} .^{3}a}}-{\frac {\operatorname {Sin} .x}{\operatorname {Log} .^{4}a}}+{\frac {\operatorname {Cos} .x}{\operatorname {Log} .^{5}a}}+\ldots \right)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9d0ffa9b475e1bc3d6cf456d33b99282f524ef84)
comme on le trouverait d’ailleurs en intégrant par partie.
Comme il s’agit toujours ici de logarithmes népériens, si l’on change
et
, il viendra simplement
![{\displaystyle \int e^{x}\operatorname {Sin} .x\operatorname {d} x=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2b5bd55bf6448831d00987e945ffa03f2434c192)
![{\displaystyle e^{x}\left(\operatorname {Sin} .x-\operatorname {Cos} .x-\operatorname {Sin} .x+\operatorname {Cos} .x+\operatorname {Sin} .x-\operatorname {Cos} .x-\ldots \right),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fe9fb38fd45427538d3975b358cdcdfcba2a14c1)
![{\displaystyle \int e^{x}\operatorname {Cos} .x\operatorname {d} x=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aac83601a07360dbf67a34e4d95cc11219be3467)
![{\displaystyle e^{x}\left(\operatorname {Cos} .x+\operatorname {Sin} .x-\operatorname {Cos} .x-\operatorname {Sin} .x+\operatorname {Cos} .x+\operatorname {Sin} .x-\ldots \right).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ea8fe1645dd73fe956434b443ade8b7f8a989ead)
Pour savoir ce que valent ces séries, nous remarquerons que, suivant que le nombre des termes qu’on y admet est de l’une des formes
la première se réduit à
![{\displaystyle 0,\quad +\operatorname {Sin} .x,\quad +\operatorname {Sin} .x-\operatorname {Cos} .x,\quad -\operatorname {Cos} .x,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/922f47566903a84bdd220d15aa29926ad41f7d27)
et la seconde à
![{\displaystyle 0,\quad +\operatorname {Cos} .x,\quad +\operatorname {Cos} .x+\operatorname {Sin} .x,\quad +\operatorname {Sin} .x\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/efe0902913dd47aec11090b99fa43e4281fb845c)
prenant donc dans chacune le quart de la somme de ces quatre résultats, nous aurons
![{\displaystyle \int e^{x}\operatorname {Sin} .x\operatorname {d} x={\frac {1}{2}}e^{x}(\operatorname {Sin} .x-\operatorname {Cos} .x),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/efa0834d4e05c1a61864e0b3a5c2337d470ffd6d)
![{\displaystyle \int e^{x}\operatorname {Cos} .x\operatorname {d} x={\frac {1}{2}}e^{x}(\operatorname {Sin} .x+\operatorname {Cos} .x)\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/03e76f324ba350f2fc5c688ad258454add054a2f)
ce que la différentiation confirme parfaitement.