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

468 Chapter 14 Techniques of integration
3 Find the area betweeny =
axisfromx=0tox=1.
4 Usepartialfractions to find
∫ 3 3x+2
(a)
2 x 2 −1 dx
4x+7
4x 2 and thex
+8x+3
(b)
(c)
(d)
∫
t +3
t 2 +2t+1 dt
∫ 2t 2 +3t+1
t 3 dt
+t
∫
6t+3
2t 2 −5t+2 dt
Solutions
1 (a) 3lnx−2ln(x+1)+c
(b) 2ln(t+1)−ln(t−1)+c
1
(c)
2 ln(2x +3) + 3 ln(2x +1) +c
2
t(t +1)
(d) ln(t +1) + +c
2
(e) 2ln(x +1) − 1 x +c
2 (a) 2.1587
(b) 1.7918
(c) −0.09971
(d) 0.1823
(e) 0.9769
3 1.2456
4 (a) 1.8767
(b) ln(t +1) − 2
t +1 +c
(c) 3tan −1 t + 1 2 ln(t2 +1)+lnt+c
(d) 5ln(t−2)−2ln(2t−1)+c
REVIEWEXERCISES14
1 Usethe given substitutionto findthe following
integrals:
∫ 1
(a) (9t+2) 10 dt z=9t+2
0
∫ 5
(b) (−t+1) 6 dt z=−t+1
3
∫ 3
(c) (4x−1) 27 dx z=4x−1
∫6
√3t+1dt
(d) z=3t+1
∫
(e) (9y−2) 17 dy z=9y−2
∫ 2 3
(f) dz
0 (2z +5) 6
∫
y=2z+5
(g) t 2 sin(t 3 )dt z =t 3
∫
(h) x 2 e x3 +1 dx z =x 3 +1
∫ 0.5
(i) sin(2t)e cos(2t) dt z = cos(2t)
∫0
π
(j) sintcos 2 tdt z=cost
∫0
(k) cost √ sintdt z=sint
2 Useintegration byparts to find
∫ π/2
∫ π/2
(a) e 2x cosxdx (b) e 2x sinxdx
0
0
3 Find
∫ lnt
dt
t
using
(a)integrationby parts
(b)the substitutionz = lnt.
4 TheintegralI n isgiven by
∫ π/2
I n = sin n θ dθ
0
(a) StateI n−2 .
(b) Show
I n = n −1
n I n−2
(c) EvaluateI 0 ,I 1 ,I 2 andI 3 .
5 Evaluate
∫
t 2
(a)
t 3 +1 dt