F.5 DEFINITE INTEGRATION 1條

a) cos 2x/(sin2 x cos2 x) = 4 cos 2x/(4 sin2 x cos2 x)

= 4 cos 2x/sin2 2x

= 4 (1/sin 2x) (cos 2x/sin 2x)

= 4 csc 2x cot 2x

b) ∫(x = π/6 → π/3) cos 2x/(sin2 x cos2 x) dx

= ∫(x = π/6 → π/3) 4 csc 2x cot 2x dx

= - 4 [csc 2x] (x = π/6 → π/3)

= [csc 2x] (x = π/3 → π/6)

= csc π/3 - csc 2π/3

= 4/√3 or (4√3)/3

2012-02-04 23:56:42 補充：
b) Ans should be

4(csc π/3 - csc 2π/3)

= 4(2/√3 - 2/√3)

= 0

Q) F.5 DEFINITE INTEGRATION 1條圖中的27題

希望有師兄能解答一下!謝謝

http://imageshack.us/photo/my-images/804/img1699k.jpg/ A)a) cos 2x/(sin2 x cos2 x) = 4 cos 2x/(4 sin2 x cos2 x)

= 4 cos 2x/sin2 2x

= 4 (1/sin 2x) (cos 2x/sin 2x)

= 4 csc 2x cot 2x

b) ∫(x = π/6 → π/3) cos 2x/(sin2 x cos2 x) dx

= ∫(x = π/6 → π/3) 4 csc 2x cot 2x dx

= - 4 [csc 2x] (x = π/6 → π/3)

= [csc 2x] (x = π/3 → π/6)

= csc π/3 - csc 2π/3

= 4/√3 or (4√3)/3 b) Ans should be

4(csc π/3 - csc 2π/3)

= 4(2/√3 - 2/√3)

= 0
希望幫到你

part b) should be :

∫(x = π/6 → π/3) cos 2x/(sin2 x cos2 x) dx

= ∫(x = π/6 → π/3) 4 csc 2x cot 2x dx

= ∫(x = π/6 → π/3) 2 csc 2x cot 2x d2x

= - 2[csc 2x] (x = π/6 → π/3)

= 2[csc 2x] (x = π/3 → π/6)

= 2[csc 2π/3 - csc 2π/6]

= 2[2/√3 - 2/√3]

= 0 ..... ans