نريد حساب: I = ∫_0^(π/2)▒[(sin 2x.dx)/(1+2 sin x)] احسب: J = ∫_0^(π/2)▒[(cos x)/(1+2sin x)]
- رياضيات
- 2021-09-27
- HalaHamid
الأجوبة
J = ∫_0^(π/2)▒[(cos x)/(1+2sin x)]dx
J = 1/2*∫_0^(π/2)▒[((1+2sin x)')/(1+2sin x)]dx
J = 1/2*〖[Ln (1+2sin x)]〗_0^(π/2)
J = ½ * (Ln 3 – Ln 1) = ½ Ln 3
I + J = ∫_0^(π/2)▒(sin 2x.dx)/(1+2sin x)+∫_0^(π/2)▒(cos x.dx)/(1+2sin x)
I + J = ∫_0^(π/2)▒〖(sin 2x+cos x)/(1+2sin x) dx〗
I + J = ∫_0^(π/2)▒〖((2sin x+1)cos x)/(1+2sin x) dx〗= ∫_0^(π/2)▒〖cos x.dx〗
I + J = 〖[sin x]〗_0^(π/2) = 1
I = 1 – J = 1 – ½ Ln 3
I = 1 – Ln √3
I = Ln (e/√3)
القوائم الدراسية التي ينتمي لها السؤال