Question

求此不定积分

Answer 1

∫ 1/(sinx)^2(cosx)^2 dx
=∫ 4 /(2sinxcosx)^2 dx
=∫ 4/(sin2x)^2 dx
=2∫ 1/(sin2x)^2 d(2x)
= -2cot 2x +C,C为常数

追答

2. 令x = 2sinθ,dx = 2cosθdθ
∫ x²/√(4 - x²) dx
= ∫ 4sin²θ/√(4 - 4sin²θ) * (2cosθdθ)
= ∫ 4sin²θ/|2cosθ| * (2cosθdθ)
= 4∫ sin²θ dθ
= 2∫ (1 - cos2θ) dθ
= 2θ - 2(1/2)sin2θ + C
= 2θ - 2sinθcosθ + C
= 2arcsin(x/2) - 2(x/2)√(4 - x²)/2 + C
= 2arcsin(x/2) - (x/2)√(4 - x²) + C

Answer 2

(7)
∫dx/(sinx.cosx)^2
=4∫dx/ (sin2x)^2
=4∫(csc2x)^2 dx
=-2cot2x + C
(8)
let
x= 2sinu
dx= 2cosu du
∫x^2/√(4-x^2) dx
=4∫ (sinu)^2 du
=2∫ (1-cos2u) du
=2[ u - (1/2)sin2u ] +C
=2[ arcsin(x/2) - (1/4)x.√(4-x^2) ] +C