394 Higher Engineering Mathematics

Now try the following exercise

Exercise 153 Further problems on

integration using algebraic substitutions

In Problems 1 to 6, integrate with respect to the

variable.

1. 2sin( 4 x+ 9 )

[

−

1

2

cos( 4 x+ 9 )+c

]

2. 3cos( 2 θ− 5 )

[

3

2

sin( 2 θ− 5 )+c

]

3. 4sec^2 ( 3 t+ 1 )

[

4

3

tan( 3 t+ 1 )+c

]

4.

1

2

( 5 x− 3 )^6

[

1

70

( 5 x− 3 )^7 +c

]

5.

− 3

( 2 x− 1 )

[

−

3

2

ln( 2 x− 1 )+c

]

6. 3e^3 θ+^5 [e^3 θ+^5 +c]

In Problems 7 to 10, evaluate the definite integrals

correct to 4 significant figures.

7.

∫ 1

0

( 3 x+ 1 )^5 dx [227.5]

8.

∫ 2

0

x

√

( 2 x^2 + 1 )dx [4.333]

9.

∫ π

3

0

2sin

(

3 t+

π

4

)

dt [0.9428]

10.

∫ 1

0

3cos( 4 x− 3 )dx [0.7369]

39.4 Further worked problems on

integration using algebraic

substitutions

Problem 7. Find

∫

x

2 + 3 x^2

dx.

Letu= 2 + 3 x^2 then

du

dx

= 6 xand dx=

du

6 x

Hence

∫

x

2 + 3 x^2

dx=

∫

x

u

du

6 x

=

1

6

∫

1

u

du,

by cancelling

=

1

6

lnu+c=

1

6

ln(2+ 3 x^2 )+c

Problem 8. Determine

∫

2 x

√

( 4 x^2 − 1 )

dx.

Letu= 4 x^2 −1then

du

dx

= 8 xand dx=

du

8 x

Hence

∫

2 x

√

( 4 x^2 − 1 )

dx=

∫

2 x

√

u

du

8 x

=

1

4

∫

1

√

u

du,by cancelling

=

1

4

∫

u

− 1

(^2) du=
1
4
⎡
⎢
⎣
u
(− 1
2
)

• 1
  −
  1
  2


• 
  

  1
  ⎤
  ⎥
  ⎦+c

  
  

  1
  4
  ⎡
  ⎢
  ⎣
  u
  1
  2
  1
  2
  ⎤
  ⎥
  ⎦+c=
  1
  2
  √
  u+c

  
  

  1
  2
  √
  (4x^2 −1)+c
  Problem 9. Show that
  ∫
  tanθdθ=ln(secθ)+c.
  ∫
  tanθdθ=
  ∫
  sinθ
  cosθ
  dθ.Letu=cosθ
  then
  du
  dθ
  =−sinθand dθ=
  −du
  sinθ
  Hence
  ∫
  sinθ
  cosθ
  dθ=
  ∫
  sinθ
  u
  (
  −du
  sinθ
  )
  =−
  ∫
  1
  u
  du=−lnu+c
  =−ln(cosθ)+c=ln(cosθ)−^1 +c,
  by the laws of logarithms.