Higher Engineering Mathematics

368 INTEGRAL CALCULUS

Table 37.1 Standard 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=5 andn=2 then

∫

5 x^2 dx=

5 x^2 +^1

2 + 1

+c=

5 x^3

3

+c

(b) Whena=2 andn=3 then

∫

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 differenti-

ating 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

(b) Rearranging

∫

(1−t)^2 dtgives:

∫

(1− 2 t+t^2 )dt=t−

2 t^1 +^1

1 + 1

+

t^2 +^1

2 + 1

+c

=t−

2 t^2

2

+

t^3

3

+c

=t−t^2 +

1

3

t^3 +c

This problem shows that functions often have to be

rearranged into the standard form of

∫

axndxbefore

it is possible to integrate them.