Standard integration 369
For example,
∫
( 3 x+ 2 x^2 − 5 )dx
=
∫
3 xdx+
∫
2 x^2 dx−
∫
5dx
=
3 x^2
2
+
2 x^3
3
− 5 x+c
37.3 Standard integrals
Since integration is the reverse process of differentia-
tion thestandard integralslisted in Table 37.1 may be
deduced and readily checked by differentiation.
Table 37.1Standard integrals
(i)
∫
axndx=
axn+^1
n+ 1
+c
(except whenn=−1)
(ii)
∫
cosaxdx=
1
a
sinax+c
(iii)
∫
sinaxdx=−
1
a
cosax+c
(iv)
∫
sec^2 axdx=
1
a
tanax+c
(v)
∫
cosec^2 axdx=−
1
a
cotax+c
(vi)
∫
cosecaxcotaxdx=−
1
a
cosecax+c
(vii)
∫
secaxtanaxdx=
1
a
secax+c
(viii)
∫
eaxdx=
1
a
eax+c
(ix)
∫
1
x
dx=lnx+c
Problem 1. Determine (a)
∫
5 x^2 dx(b)
∫
2 t^3 dt.
The standard integral,
∫
axndx=
axn+^1
n+ 1
+c
(a) Whena=5andn=2then
∫
5 x^2 dx=
5 x^2 +^1
2 + 1
+c=
5 x^3
3
+c
(b) Whena=2andn=3then
∫
2 t^3 dt=
2 t^3 +^1
3 + 1
+c=
2 t^4
4
+c=
1
2
t^4 +c
Each of these results may be checked by differentiating
them.
Problem 2. Determine
∫(
4 +
3
7
x− 6 x^2
)
dx.
∫
( 4 +^37 x− 6 x^2 )dxmay be written as
∫
4dx+
∫ 3
7 xdx−
∫
6 x^2 dx, i.e. each term is integrated
separately. (This splitting up of terms only applies,
however, for addition and subtraction.)
Hence
∫(
4 +
3
7
x− 6 x^2
)
dx
= 4 x+
(
3
7
)
x^1 +^1
1 + 1
−( 6 )
x^2 +^1
2 + 1
+c
= 4 x+
(
3
7
)
x^2
2
−( 6 )
x^3
3
+c
= 4 x+
3
14
x^2 − 2 x^3 +c
Note that when an integral contains more than one term
there is no need to have an arbitrary constant for each;
just a single constant at the end is sufficient.
Problem 3. Determine
(a)
∫
2 x^3 − 3 x
4 x
dx (b)
∫
( 1 −t)^2 dt.
(a) Rearranging into standard integral form gives:
∫
2 x^3 − 3 x
4 x
dx
=
∫
2 x^3
4 x
−
3 x
4 x
dx=
∫
x^2
2
−
3
4
dx
=
(
1
2
)
x^2 +^1
2 + 1
−
3
4
x+c
=
(
1
2
)
x^3
3
−
3
4
x+c=
1
6
x^3 −
3
4
x+c