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

Solution (a)
sinx+cosx
(j)
2
(k) 2t −e t
∫
∫
x 2 +9dx=
∫
x 2 dx+
( ) z −1
(l) tan
2
(m) e t +e −t
9 dx usinglinearity
13.2 Elementary integration 433
(n) 3sec(4x −1)
(o) 2cot9x
(p) 7cosec(π/3)
(b)
+9x +c usingTable 13.1
3
Note thatonly a single constant of integration isrequired.
∫
3t 4 − √ ∫ ∫
tdt=3 t 4 dt − t 1/2 dt using linearity
= x3
( ) t
5
=3 − t3/2
+c using Table 13.1
5 3/2
= 3t5
5 − 2t3/2
+c
3
(c)
∫ 1
dx = ln |x| +c using Table 13.1.
x
Sometimesitisconvenienttousethelawsoflogarithmstorewriteanswersinvolving
logarithms. For example, we can write ln|x| +c as ln|x| +ln|A| wherec = ln |A|.
This enables us towritethe integralas
∫ 1
dx = ln|Ax|
x
(d)
∫
∫
(t+2) 2 dt=
t 2 +4t+4dt= t3 3 +2t2 +4t+c
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
∫ 1
z +zdz=ln|z|+z2 2 +c
∫
4e 2z dz = 4e2z
2 +c=2e2z +c
∫
3sin(4t)dt = − 3cos4t +c
4
∫
4sin(9x +2)
4cos(9x +2)dx = +c
9
∫
3e 2z dz = 3e2z
2 +c
∫ sinx+cosx
dx = −cosx+sinx +c
2 2
∫
2t−e t dt=t 2 −e t +c
∫ ( )
( )∣ z −1
tan dz=2ln
z −1 ∣∣∣
2 ∣ sec +c
2