Integration using a

table of
anti-derivatives
mc-TY-inttable-2009-1

We may regard integration as the reverse of differentiation. So if we have a table of derivatives,

we can read it backwards as a table of anti-derivatives. When we do this, we often need to deal
with constants which arise in the process of differentiation.

In order to master the techniques explained here it is vital that you undertake plenty of practice
exercises so that they become second nature.

After reading this text, and/or viewing the video tutorial on this topic, you should be able to:

• use a table of derivatives, or a table of anti-derivatives, in order to integrate simple func-

tions.

Contents
1. Introduction 2
2. Integrating powers 3
3. Integrating exponentials 3
4. Integrating trigonmetric functions 4
5. Integrals giving rise to inverse trigonometric functions 5

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1. Introduction
When we are integrating, we need to be able to recognise standard forms. The following table
gives a list of standard forms, obtained as anti-derivatives. Sometimes, it may be possible to
use one of these standard forms directly. On other occasions, some manipulation will be needed
first.

Key Point

xn+1
Z
xn dx = +c (n 6= −1)
n+1
(ax + b)n+1
Z
(ax + b)n dx = +c (n 6= −1)
a(n + 1)
1
Z
dx = ln |x| + c
x
1 1
Z
dx = ln |ax + b| + c
ax + b a
Z
ex dx = ex + c
1 mx
Z
emx dx = e +c
m
Z
cos x dx = sin x + c
1
Z
cos nx dx = sin nx + c
n
Z
sin x dx = − cos nx + c
1
Z
sin nx dx = − cos nx + c
n
Z
sec2 x dx = tan x + c
1
Z
sec2 nx dx = tan nx + c
n
1
Z
√ dx = sin−1 x + c
1−x 2

1 x
Z
√ dx = sin−1 +c
a 2 − x2 a
1
Z
dx = tan−1 x + c
1 + x2
1 1 −1 x
Z  
dx = tan +c
a2 + x2 a a

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2. Integrating powers
We know that the derivative of xn is nxn−1 . Replacing n by n + 1 we see that the derivative of
n+1 n xn+1
x is (n + 1)x , so that the derivative of is xn (provided that n + 1 6= 0). Thus
n+1
xn+1
Z
xn dx = +c.
n+1

Similarly, the derivative of (ax + b)n is an(ax + b)n−1 . Replacing n by n + 1 we see that the
(ax + b)n+1
derivative of (ax + b)n+1 is a(n + 1)(ax + b)n , so that the derivative of is (ax + b)n
a(n + 1)
(provided that n + 1 6= 0 and that a 6= 0). Thus

(ax + b)n+1
Z
(ax + b)n dx = +c.
a(n + 1)

What happens if n = −1, so that n + 1 = 0? We know that the derivative of ln |x| is 1/x, so
that
1
Z
dx = ln |x| + c .
x
a 1 1
Simlarly, the derivative of ln |ax + b| is , so that the derivative of ln |ax + b| is .
ax + b a ax + b
Thus
1 1
Z
dx = ln |ax + b| + c .
ax + b a
Example
1
Z
Find dx.
2 − 3x
Here, a = −3 and b = 2, so
1 1
Z
dx = − ln |2 − 3x| + c .
2 − 3x 3

3. Integrating exponentials
We know that the derivative of ex remains unchanged, as ex . Thus
Z
ex dx = ex + c .

1
Similarly, we know that the derivative of emx is memx , so that the derivative of emx is emx .
m
Thus
1
Z
emx dx = emx + c .
m

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Example
Z
Find e4x dx.

Here, m = 4, so
1
Z
e4x dx = e4x + c .
4

4. Integrating trigonometric functions

We know that the derivative of sin x is cos x. Thus
Z
cos x dx = sin x + c .

1
Similarly, we know that the derivative of sin nx is n cos nx, so that the derivative of sin nx is
n
cos nx. Thus
1
Z
cos nx dx = sin nx + c .
n
We also know that the derivative of cos x is − sin x. Thus
Z
sin x dx = − cos x + c .

1
Similarly, we know that the derivative of cos nx is −n sin nx, so that the derivative of − cos nx
n
is sin nx. Thus
1
Z
sin nx dx = − cos nx + c .
n
sin x
We can use the fact that tan x = to find an anti-derivative of tan x. We use the rule for
cos x
− sin x
logarithmic differentiation to see that the derivative of ln | cos x| is , so that
cos x
sin x
Z Z
tan x dx = dx
cos x
= − ln | cos x| + c
= ln | sec x| + c .
(In the last step of this argument, we have used the fact that − ln u is equal to ln(1/u).)
There is one more trigonometric function which we can integrate without difficulty. We know
that the derivative of tan x is sec2 x. Thus
Z
sec2 x dx = tan x + c .

1
Similarly, the derivative of tan nx is n sec2 nx, so that the derivative of tan nx is sec2 nx. Thus
n
1
Z
sec2 nx dx = tan nx + c
n

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5. Integrals giving rise to inverse trigonometric functions
Sometimes, integrals involving fractions and square roots give rise to inverse trigonometric func-
tions.
1
We know that the derivative of sin−1 x is √ . Thus
1 − x2

1
Z
√ dx = sin−1 x + c .
1 − x2
x 1 1
Similarly, we know that the derivative of sin−1 is q , which equals √ . Thus
a x 2

a2 − x2
a 1− a

1 x
Z
√ dx = sin−1 +c.
a2 − x2 a

Example
1
Z
Find √ dx.
4 − x2
√
Here, a = 4 = 2, so that
1 x
Z
√ dx = sin−1 +c.
4 − x2 2

Example
1
Z
Find √ dx.
4 − 9x2

This is not quite in our standard form. However, we can take the 9 outside the square root, so
that it becomes 3. We get

1 1 1
Z Z
√ dx = ·q dx ,
4 − 9x2 3 4
− x2
9

1
and this is in the standard form. So now we can take the 3
outside the integral, and we see that
q
a = 49 = 32 , so that

1 1 1
Z Z
√ dx = q dx
4 − 9x2 3 − x2 4
9
 
1 −1 x
= sin 2 +c
3 3
 
1 −1 3x
= sin +c.
3 2

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Another type of integral which may be found using an inverse trigonometric function involves a
fraction, but does not involve a square root.
1
We know that the derivative of tan−1 x is . Thus
1 + x2
1
Z
dx = tan−1 x + c .
1 + x2
x 1 a
Similarly, we know that the derivative of tan−1 is 2 , which equals 2 , so
a a(1 + xa ) a + x2


1 x 1
that the derivative of tan−1 is 2 . Thus
a a a + x2
1 1 −1 x
Z  
dx = tan +c
a2 + x2 a a
Example
1
Z
Find dx.
9 + x2
√
Here, a = 9 = 3, so that
1 1 −1 x
Z  
dx = tan +c.
9 + x2 3 3
Example
1
Z
Find dx.
25 + 16x2
Here, we take the 16 outside the integral, so that we get
1 1 1
Z Z
dx = 25 dx .
25 + 16x2 16 16
+ x2

q
Now we can see that a = 25 16
= 54 , so that
!
1 1 1 x
Z
dx = × 5
tan −1
5

+c
25 + 16x2 16 4 4
 
1 4 4x
= × tan−1 +c
16 5 5
 
1 4x
= tan −1
+c.
20 5

Exercises
1. Determine the integral of each of the following functions
1 1 1
(a) x8 (b) (c) √ (d) (e) sin 5x
x3 x x
1 1 1 1
(f) sec2 2x (g) (h) √ (i) √ (j)
16 + x2 4 − x2 16 − 9x2 4 + 25x2

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2. Integration has the same linearity rules as differentiation, namely
Z Z Z Z Z
kf (x) dx = k f (x) dx and f (x) + g(x) dx = f (x) dx + g(x) dx .

Use these rules to determine the integrals of the following functions

√
(a) 5x4 + 10 cos 2x (b) 12e4x + 4 x (c) 36 sec2 4x + 12e−3x
54 15
(d) 12x6 − 2 sin 4x (e) +√ (f) 10 cos 5x − 5 cos 10x
9+x 2
9 − x2
Answers
1. In all answers the constant of integration has been omitted.
1 9 1 1 √
(a) x (b) − x−2 = − 2 (c) 2x1/2 = 2 x (d) ln x
9 2 2x
1 1 1 −1 x
  x
(e) − cos 5x (f) tan 2x (g) tan (h) sin −1
5 2 4 4 2
   
1 −1 3x 1 5x
(i) sin (j) tan −1
3 4 10 2

2. In all answers the constant of integration has been omitted.

8
(a) x5 + 5 sin 2x (b) 3e4x + x3/2 (c) 9 tan 4x − 4e−3x
3
12 7 1 −1 x
 
−1 x
  1
(d) x + cos 4x (e) 18 tan + 15 sin (f) 2 sin 5x − sin 10x
7 2 3 3 2

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