082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

14.3 Integration by substitution 463
3 (a) 0.1006 (b) 0.7854 (c) 1.9488
4 (a)
(d) 6.9986 (e) 0.4542
(b)
e 2t (2t 2 −2t +1)
4
+c
)
2
(t 2
9 tcos3t+ 3 − 2 sin3t+c
27
( ) ( )
t
(c) 8tsin −2(t 2 t
−8)cos +c
2 2
5 (a) 12.7781 (b) 0 (c) −0.1834
6 I n = tn e kt
− n k k I n−1 , e3t (9t 2 −6t +2)
,
27
e 3t (9t 3 −9t 2 +6t −2)
,
27
e 3t (27t 4 −36t 3 +36t 2 −24t +8)
81
7 I n =n
8 4.2281
{(
π
2
) n−1
− (n −1)I n−2
}
1.4022,2.3963,4.5084
14.3 INTEGRATIONBYSUBSTITUTION
∫
Example14.6 Find
This technique is the integral equivalent of the chain rule. It is best illustrated by
examples.
(3x +1) 2.7 dx.
Solution Letz = 3x + 1, so that dz
dx
becomes
∫
z 2.71 3 dz = 1 ∫
z 2.7 dz = 1 ( z
3.7
3 3 3.7
dz
= 3, that is dx = . Writing the integral in terms ofz, it
3
)
+c = 1 (3x +1) 3.7
+c
3 3.7
Example14.7 Evaluate
∫ 3
Solution Let v =t 2 so dv
dt
2
dt = 1 2t dv
tsin(t 2 )dt.
= 2t, thatis
When changing the integral from one in terms oft to one in terms of v, the limits must
alsobe changed. Whent = 2, v = 4;whent = 3, v = 9.Hence, the integral becomes
∫ 9
4
sinv
2 dv = 1 2 [−cos v]9 4 = 1 [−cos9 +cos4] = 0.129
2
Sometimesthe substitution can involve a trigonometric function.