INTEGRATION USING ALGEBRAIC SUBSTITUTIONS 393

H

7.

∫ 1

0

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

8.

∫ 2

0

x

√

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

9.

∫ π

3

0

2 sin

(

3 t+

π

4

)

dt [0.9428]

10.

∫ 1

0

3 cos (4x−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

√

(4x^2 −1)

dx.

Letu= 4 x^2 −1 then

du

dx

= 8 xand dx=

du

8 x

Hence

∫

2 x

√

(4x^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
  Hence
  ∫
  tanθdθ=ln(secθ)+c,
  since (cosθ)−^1 =
  1
  cosθ
  =secθ
  39.5 Change of limits
  When evaluating definite integrals involving substi-
  tutions it is sometimes more convenient tochange
  the limitsof the integral as shown in Problems 10
  and 11.
  Problem 10. Evaluate
  ∫ 3
  15 x
  √
  (2x^2 +7) dx,
  taking positive values of square roots only.
  Letu= 2 x^2 +7, then
  du
  dx
  = 4 xand dx=
  du
  4 x