tolong bantu jelaskan, dimana dpt integral sec³teta tan teta menjadi ⅓ sec³ teta (yg dikotakkan ) terima kasih

[tex]\displaystyle \int\sec^3\theta\tan\theta-\tan\theta\sec\theta\,d\theta=\int\sec^2\theta\sec\theta\tan\theta-\tan\theta\sec\theta\,d\theta\\\int\sec^3\theta\tan\theta-\tan\theta\sec\theta\,d\theta=\int\frac33\sec^2\theta\sec\theta\tan\theta-\tan\theta\sec\theta\,d\theta\\\int\sec^3\theta\tan\theta-\tan\theta\sec\theta\,d\theta=\int\frac13\cdot3\sec^2\theta\sec\theta\tan\theta-\tan\theta\sec\theta\,d\theta\\\int\sec^3\theta\tan\theta-\tan\theta\sec\theta\,d\theta=\frac13\sec^3\theta-\sec\theta+C[/tex]


[tex]\displaystyle \text{perhatikan turunan berikut :}\\\text{misal :}\\\sec\theta=u\\\\\frac{d(\sec^3\theta)}{d\theta}=\frac{d(u^3)}{du}\cdot\frac{du}{d\theta}\\\frac{d(\sec^3\theta)}{d\theta}=\frac{d(u^3)}{du}\cdot\frac{d(\sec\theta)}{d\theta}\\\frac{d(\sec^3\theta)}{d\theta}=3u^2\cdot\sec\theta\tan\theta\\\frac{d(\sec^3\theta)}{d\theta}=3\sec^2\theta\cdot\sec\theta\tan\theta\\\frac{d(\sec^3\theta)}{d\theta}=3\sec^3\theta\tan\theta[/tex]

INTEGRAL TRIGONOMETRI
∫ sec^2 x dx = Tan x + C
∫ Tan x sec x dx = sec x + C

TURUNAN TRIGONOMETRI
y = Tan x => dy = sec^2 x dx
y = sec x => dy = Tan x sec x dx

Misal
u = sec x
du = Tan x sec x dx

∫ sec^3 x Tan x dx
= ∫ sec^2 x . sec x Tan x dx
= ∫ u^2 du
= 1/3 u^3 + C
= 1/3 sec^3 x + C

^ = pangkat