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2 置換積分法 20140417

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4STEP 数学Ⅲ(新課程)を解いてみた
積分法 2
http://toitemita.sakura.ne.jp
置換積分法
コツ
できるだけ大きい式のかたまりを置換すると式処理が楽。
377
(1)
3
1 + x = t とおくと, 1 + x = t 3 より,
dx
= 3t 2
dt
\ dx = 3t 2 dt
また, x = t 3 - 1
ò (t
= 3ò (t
ò
\ x3 1 + x dx =
)
3
- 1 t × 3t 2 dt
6
- t 3 dt
)
1 ö
æ1
= 3ç t 7 - t 4 ÷ + C
4 ø
è7
3 4 3
=
t 4t - 7 + C
28
3
=
(1 + x )3 1 + x {4(1 + x ) - 7} + C
28
3
=
(4 x - 3)(1 + x )3 1 + x + C
28
(
)
(2)
cos x = t とおくと, - sin x =
ò
ò
= -ò t
dt
より, sin xdx = - dt
dx
\ sin x cos 4 xdx = cos 4 x sin xdx
4
dt
1
= - t5 + C
5
1
= - cos 5 x + C
5
(3)
tan x = t とおくと, 1 + t 2 =
\
dx
ò cos
4
2
cos x
,
1
2
cos x
=
dx
1
x ò cos x cos
= ò (1 + t )dt
=
1
2
2
x
2
1 3
t +C
3
1
= tan x + tan 3 x + C
3
=t+
1
dt
dx
= dt
より,
dx
cos 2 x
4STEP 数学Ⅲ(新課程)を解いてみた
http://toitemita.sakura.ne.jp
(4)
ex
\
2
+ x +5
= t とおくと, (2 x + 1)e x
ò (2 x + 1)e
x 2 + x +5
2
+ x+5
=
2
dt
より, (2 x + 1)e x + x + 5 dx = dt
dx
ò
dx = dt
=t+C
= ex
2
+ x +5
+C
(5)
e x + 2 = t とおくと, e x dx = dt
\
ò (e
e 2x
x
+2
)
2
dx =
ò (e
ex
)
2
x
e x dx
+2
t-2
=
dt
t2
æ1
ö
= ç - 2t - 2 ÷dt
èt
ø
ò
ò
= log t + 2t -1 + C
(
)
= log e x + 2 +
2
x
e +2
+C
(6)
log x - 1 = t とおくと,
\
log x
ò x(log x - 1)
2
dx =
=
dx
= dt
x
log x
ò (log x - 1)
ò
2
dx
x
t +1
dt
t2
æ1
ö
= ç + t - 2 ÷dt
èt
ø
1
= log t - + C
t
ò
= log log x - 1 -
1
+C
log x - 1
2
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