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

14.3 Integration by substitution 465
The resultofExample 14.11 isparticularly important.
∫ df/dx
dx=ln|f|+c
f
∫ 4
3t
Example14.12 2 +2t
Evaluate
2 t 3 +t 2 +1 dt.
Solution The numerator isthe derivative ofthe denominator and so
∫ 4
2
3t 2 +2t
t 3 +t 2 +1 dt = [ln|t3 +t 2 +1|] 4 2 =ln81−ln13=1.83
Example14.13 Find
∫
(a)
∫
(b)
∫
(c)
4
5x−7 dx
t
t 2 +1 dt
e t/2
e t/2 +1 dt
Solution The integrands arerewritten so thatthe numerator isthe derivative of the denominator.
∫
4
(a)
5x−7 dx = 4 ∫
5
5 5x−7 dx = 4 5 ln|5x−7|+c
∫
t
(b)
t 2 +1 dt = 1 ∫
2t
2 t 2 +1 dt
(c)
∫
= 1 2 ln|t2 +1|+c
e t/2 ∫ 1
e t/2 +1 dt=2 2 et/2
e t/2 +1 dt =2ln|et/2 +1|+c
EXERCISES14.3
1 Usethe given substitutionsto findthe following
integrals:
∫
(a) (4x+1) 7 dx, z=4x+1
∫
(b) t 2 sin(t 3 +1)dt, z =t 3 +1
∫
(c) 4te −t2 dt, z=t 2
∫
(d) (1−z) 1/3 dz, t=1−z
∫
(e) cost(sin 5 t)dt, z =sint
2 Evaluate the following definiteintegrals:
∫ 2
∫ π/2
(a) (2t +3) 7 dt (b) sin2tcos 4 2tdt
1
0
∫ 1
∫ 2
(c) 3t 2 √
e t3 dt (d) 4+3xdx
0
0
(
∫ √x )
2 sin
(e) √ dx
1 x
3 Find the area betweeny =x(3x 2 +2) 4 andthexaxis
fromx=0tox=1.