Question

根号（1+sin2x) dx 的不定积分怎么求呢？

Answer 1

J = ∫ √(1+sin2x) dx
= (1/2)∫ √(1+sint) dt，t=2x
Let y = 1+sint then dy = costdt = √y√(2-y)dt
J = ∫ √y * 1/[√y√(2-y)] dy
= ∫ 1/√(2-y) dy
= -∫ d(2-y)/√(2-y)
= -√(2-y) + C
= -√[2-(1+sin2x)] + C
= -√(1-sin2x) + C

∫[π/6，π/3] (1+cotx)² dx
= ∫[π/6，π/3] (1+2cotx+cot²x) dx
= ∫[π/6，π/3] (1+2cotx) dx + ∫[π/6，π/3] (csc²x-1) dx
= [2ln|sinx| - cotx] [π/6，π/3]
= [2lnsin(π/3) - cot(π/3)] - [2lnsin(π/6) - cot(π/6)]
= 2/√3 + ln(3)

Answer 2

把正弦拆开，然后把余弦放到后面积分就行了