Réponse :
∫ [1/(1+sin(x))] dx = ∫ [1/(1+sin(x))] * [(1 - sin (x))/(1 - sin(x))] dx
= ∫ [(1 - sin(x))/(1+sin(x))(1 - sin(x))] dx = ∫ [(1 - sin(x))/(1 - sin²(x))] dx
or 1 - sin²(x) = cos²(x) + sin²(x) - sin²(x) = cos²(x)
= ∫[(1 - sin(x))/cos²(x)] dx = ∫ [1/cos²(x)] dx + ∫- sin(x)/cos²(x)] dx
on pose u = cos(x)
du = - sin(x)dx or ∫1/cos²(x)dx = tan(x) + C
= tan(x) + ∫(1/u²)du = tan(x) + ∫u⁻²du = tan(x) + u⁻²⁺¹/(-2+1) = tan(x) - 1/u
= tan (x) - 1/cos(x) + C
Donc ∫ [1/(1+sin(x))] dx = tan (x) - 1/cos(x) + C
Explications étape par étape :
