作者k87 (谁不希望天长地久)
看板trans_math
标题Re: [积分] 几题积分请教
时间Tue Apr 24 02:48:48 2007
※ 引述《c1986 (可怜重考生)》之铭言：
: 3
: 1. ∫ x^2√(9-x^2)dx
: 0
∫ x^2√(9-x^2)dx 令 x = 3 sin u x = 3 => u = π/2
dx = 3 cos u du x = 0 => u = 0
= ∫ 9(sin u)^2 √(9-9(sin u)^2) 3 cos u du
= ∫ 9(sin u)^2˙3 (cos u)˙3 (cos u) du
= 81 ∫ (sin u)^2˙(cos u)^2 du
81
= --- ∫ (sin 2u)^2 du
4
81
= --- ∫ (sin 2u)^2 d 2u 令 2u = t
8
81 1 - cos 2t
= --- ∫ -------------- dt
8 2
81 1 81 cos 2t
= --- ∫ --- dt - --- ∫ --------- dt
8 2 8 2
81 81
= ---- t - ---- sin 2t + c
16 32
81 81
= ---- 2u - ---- sin 4u + c
16 32
3 π/2
∫ x^2√(9-x^2)dx = ∫ 9(sin u)^2 √(9-9(sin u)^2) 3 cos u du
0 0
81 81 |π/2
= ---- 2u - ---- sin 4u |
16 32 |0
81
= ---- π
16
: x
: 2 ∫---------- dx
: √(x^2-2x)
x x
∫----------- dx = ∫---------------- dx 令 x-1 = sec u
√(x^2-2x) √((x-1)^2 -1) dx = sec u˙tan u du
(sec u +1)˙sec u˙tan u
= ∫-------------------------- du
x-1 ◢ tan u
◢█ √((x-1)^2 -1)
1 = ∫ (sec u)^2 du + ∫ sec u du
= tan u + ln| sec u + tan u | + c
= √((x-1)^2 -1) + ln| (x-1) + √((x-1)^2 -1)| + c
--
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