Please solve this whole Integration of [sec^2xdx/sqrt (1-tan^2x)]?
回答 (1)
2016-06-14 4:18 pm
Let u = tan x
Then du = sec²x dx
∫ sec²x dx / √(1 - tan²x)
= ∫du / √(1 - u²)
Let u = sin θ
Then du = cosθ dθ and θ = arcsin u
∫ sec²x dx / √(1 - tan²x)
= ∫du / √(1 - u²)
= ∫ cosθ dθ / √(1 - sin²θ)
= ∫ cosθ dθ / √(cos²θ)
= ∫ cosθ dθ / cosθ
= ∫ dθ
= θ + C
= arcsin u + C
= arcsin(tanx) + C
Then du = sec²x dx
∫ sec²x dx / √(1 - tan²x)
= ∫du / √(1 - u²)
Let u = sin θ
Then du = cosθ dθ and θ = arcsin u
∫ sec²x dx / √(1 - tan²x)
= ∫du / √(1 - u²)
= ∫ cosθ dθ / √(1 - sin²θ)
= ∫ cosθ dθ / √(cos²θ)
= ∫ cosθ dθ / cosθ
= ∫ dθ
= θ + C
= arcsin u + C
= arcsin(tanx) + C
收錄日期: 2021-04-18 15:00:40
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