1. Theory

Consider an integral of the form:

 ∫ ƒ(ax + b)

where a and b are constants. We have here an unspecified function ƒ of a linear function x.

Letting u = ax + b, then du/dx = a, and this gives dx = du/a

This allows us to change the integration variable from x to u.

 ∫ ƒ(ax + b) dx =  ∫ ƒ(u)du/a

The final result is

 ∫ ƒ(ax + b) dx = 1/a ∫ ƒ(u) du

- where u = ux + b

This is a general result for integrating functions of a linear function x.

Each application of this result involves dividing by the coefficient of x and then integrating.

2. Exercises

Perform the following integrations.

Click on the questions to reveal their solution

Exercise 1:

 ∫ (2x − 1)3 dx

Solution: Let u = 2x − 1, then du/ dx = 2 and dx = du/2.

 ∫ (2x−1)3dx =  ∫ u3 du/2 = 1/2 ∫ u3du
  = 1/2⋅1/4 u4+C
  = 1/8 (2x − 1)4 + C

Note: The final result can also be obtained using the general pattern:

 ∫ ƒ(ax + b) dx = 1/a ∫ ƒ(u) du

- where a = 2 and  ∫ ƒ(u) du is  ∫ u3 du.

Exercise 2:

 ∫ cos (3x + 5) dx

Solution: Let u = 3x + 5, then du/dx = 3 and dx = du/3.

 ∫ cos (3x + 5) dx =  ∫ cos (u)⋅ du/3 = 1/3 ∫ cos (u) du
  = 1/3 sin (u) + C
  = 1/3 sin (3x + 5)4 + C

Note: The final result can also be obtained using the general pattern:

 ∫ ƒ(ax + b) dx = 1/a ∫ ƒ(u) du

- where a = 3 and  ∫ ƒ(u) du is  ∫ cos (u) du.

Exercise 3:

 ∫ e5x + 2 dx

Solution: Let u = 5x + 2, then du/ dx = 5 and dx =  du/5.

 ∫ e5x+2dx =  ∫ eu du/5 = 1/5 ∫ eudu
  = 1/5 eu + C
  = 1/5 e5x + 2 + C

Note: The final result can also be obtained using the general pattern:

 ∫ ƒ(ax + b) dx = 1/a ∫ ƒ(u) du

- where a = 5 and  ∫ ƒ(u) du is  ∫ eu du.

Exercise 4:

 ∫ sinh (3x) dx

Solution: Let u = 3x, then du/dx = 3 and dx = du/3.

 ∫ sinh (3x) dx =  ∫ sinh (u)⋅du/3 = 1/3 ∫ sinh (u) du
  = 1/3 cosh (u) + C
  = 1/3 cosh (3x) + C

Note: The final result can also be obtained using the general pattern:

 ∫ ƒ(ax + b) dx = 1/a ∫ ƒ(u) du

- where a = 3 and  ∫ ƒ(u) du is  ∫ sinh (u) du.

Exercise 5:

 ∫ dx2x − 1

Solution: Let u = 2x − 1, then du/dx = 2 and dx = du/2.

 ∫ dx2x − 1 =  ∫ duu⋅12 = 1/2 ∫ duu
  = 1/2 ln u + C
  = 1/2 ln 2x − 1 + C

Note: The final result can also be obtained using the general pattern:

 ∫ ƒ(ax + b) dx = 1/a ∫ ƒ(u) du

- where a = 2 and  ∫ ƒ(u) du is  ∫ du/u

Exercise 6:

 ∫ dx1 + (5x)2

Solution: Let u = 5x − 1, then du/dx = 5 and dx = du/5.

 ∫ dx1 + (5x)2 =  ∫ du1 + u2⋅15 = 1/5 ∫ du1 + u2
  = 1/5 tan−1 u + C
  = 1/5 tan−1 5x + C

Note: The final result can also be obtained using the general pattern:

 ∫ ƒ(ax + b) dx = 1/a ∫ ƒ(u) du

- where a = 5 and  ∫ ƒ(u) du is  ∫ du/1 + u2

Exercise 7:

 ∫ sec2 (7x + 1) dx

Solution: Let u = 7x + 1, then du/dx = 7 and dx = du/7.

 ∫ sec2 (7x + 1) dx = 1/7 ∫ sec2 u du
  = 1/7 tanu + C
  = 1/7 tan (7x + 1) + C

Note: The final result can also be obtained using the general pattern:

 ∫ ƒ(ax + b) dx = 1/a ∫ ƒ(u) du

- where a = 7 and  ∫ ƒ(u) du is  ∫ sec2udu.

Exercise 8:

 ∫ sin (3x − 1) dx

Solution: Let u = 3x − 1, then du/dx = 3 and dx = du/3.

 ∫ sin (3x − 1) dx = 1/3 ∫ sin (u) du
  = −1/3 cos (u) + C
  = −1/3 cos (3x − 1) + C

Note: The final result can also be obtained using the general pattern:

 ∫ ƒ(ax + b) dx = 1/a ∫ ƒ(u) du

- where a = 3 and  ∫ ƒ(u) du is  ∫ sin (u) du.

Exercise 9:

 ∫ cosh (1 + 2x) dx

Solution: Let u = 1 + 2x, then du/dx = 2 and dx = du/2.

 ∫ cosh (1 + 2x) dx = = 1/2 ∫  cosh (u) du
  = 1/2 sinh (u) + C
  = 1/2 sinh (1 + 2x) + C

Note: The final result can also be obtained using the general pattern:

 ∫ ƒ(ax + b) dx = 1/a ∫ ƒ(u) du

- where a = 2 and  ∫ ƒ(u) du is  ∫ cosh (u) du.

Exercise 10:

 ∫ tan (9x − 1) dx

Solution: Let u = 9x − 1, then du/dx = 9 and dx = du/9.

 ∫ tan (9x − 1) dx =  ∫ tan(u) du/9 = 1/9 ∫ tan (u) du
  = −1/9 ln cos (u) + C
  = −1/9 ln cos (9x − 1) + C

Note: The final result can also be obtained using the general pattern:

 ∫ ƒ(ax + b) dx = 1/a ∫ ƒ(u) du

- where a = 9 and  ∫ ƒ(u) du is  ∫ tan (u) du.

3. Standard Integrals

ƒ(x) ∫ƒ(x) dx   ƒ(x) ∫ƒ(x) dx
xn xn+1/n+1 (n ≠ −1)   [g(x)]n g'(x) [g(x)]n+1/n+1 (n ≠ −1)
1/x ln x   g'(x)/g(x) ln g(x)
ex ex   ax ax/ln a (a > 0)
sin x −cos x   sinh x cosh x
cos x sin x   cosh x sinh x
tan x − ln cosx   tanh x ln cosh x
cosec x ln tan  x/2   cosech x ln tanh  x/2
sec x ln sec x + tan x   sec x 2 tan−1ex
sec2 x tan x   sec2 x tanh x
cot x ln sin x   cot x ln sinh x
sin2 x x/2 − sin 2x/4   sinh2 x sinh 2x/4 − x/2
cos2 x x/2 + sin 2x/4   cosh2 x sinh 2x/4 + x/2
1/a2 + x2 1/a tan −1  x/a   (a > 0)   √a2 + x2 a2/2[ sinh−1(x/a) + x √a2−x2/a2]
1/a2 − x2 1/2a ln  a + x/a − x    (0 < x < a)   √a2−x2 a2/2[ sin−1(x/a) + x √a2−x2/a2]
1/x2 − a2 1/2a ln  x − a/x + a   (x > a > 0)   √x2 − a2 a2/2[−cosh−1(x/a) + x √x2−a2/a2]
1/ √ a2 + x2 ln  x + √a2 + x2/a   (a > 0)      
1/ √ a2 − x2 sin−1  x/a     (−a < x < a)   1/ √ x2 − a2 ln  x + √x2 − a2/a   (x > a > 0)

PPLATO material © copyright 2004, University of Salford