Introduction to integration 327

subtraction.) Hence,
∫(
3 +

2

5

x− 6 x^2

)

dx

= 3 x+

(

2

5

)

x^1 +^1

1 + 1

−( 6 )

x^2 +^1

2 + 1

+c

= 3 x+

(

2

5

)

x^2

2

−( 6 )

x^3

3

+c= 3 x+

1

5

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 constantcat the end is sufficient.

Problem 6. Determine

∫ (

2 x^3 − 3 x

4 x

)

dx

Rearranging into standard integral form gives
∫ (
2 x^3 − 3 x
4 x

)

dx=

∫ (

2 x^3

4 x

−

3 x

4 x

)

dx

=

∫ (

1

2

x^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

Problem 7. Determine

∫(

1 −t

) 2

dt

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 example shows that functionsoftenhave tobe rear-
ranged into the standard form of

∫
axndxbefore it is
possible to integrate them.

Problem 8. Determine

∫

5

x^2

dx

∫

5

x^2

dx=

∫

5 x−^2 dx

Using the standard integral,

∫

axndx,whena=5and

n=−2, gives

∫

5 x−^2 dx=

5 x−^2 +^1

− 2 + 1

+c=

5 x−^1

− 1

+c

=− 5 x−^1 +c=−

5

x

+c

Problem 9. Determine

∫

3

√

xdx

For fractional powers it is necessary to appreciate

√nam=amn

∫

3

√

xdx=

∫

3 x

1

(^2) dx=
3 x
1
2 +^1
1
2

• 
  

  1
  +c=
  3 x
  3
  2
  3
  2
  +c
  = 2 x
  3
  (^2) +c= 2
  √
  x^3 +c
  Problem 10. Determine
  ∫
  − 5
  9
  √ 4
  t^3
  dt
  ∫
  − 5
  9
  √ 4
  t^3
  dt=
  ∫
  − 5
  9 t
  3
  4
  dt=
  ∫ (
  −
  5
  9
  )
  t−
  3
  (^4) dt

  
  

  (
  −
  5
  9
  )
  t−
  3
  4 +^1
  −
  3
  4

  
  


• 
  

  1
  +c

  
  

  (
  −
  5
  9
  )
  t
  1
  4
  1
  4
  +c=
  (
  −
  5
  9
  )(
  4
  1
  )
  t
  1
  (^4) +c
  =−
  20
  9
  √ (^4) t+c
  Problem 11. Determine
  ∫
  4cos3xdx
  From 2 of Table 35.1,
  ∫
  4cos3xdx=( 4 )
  (
  1
  3
  )
  sin3x+c

  
  

  4
  3
  sin 3x+c
  Problem 12. Determine
  ∫
  5sin2θdθ