How to approach an integration
Date- 10/13/04
Instructor – Dibyajyoti Deb
Whatever I am writing here is what I think about the subject and what I would have done if I were
given these integrations. So my views are bound to be different from any other views and in some
cases may be different from the book. There may be other easier methods to solve depending upon the
problem.
Whenever you are dealing with an integration your main aim should be to reduce the problem to any
simpler form which should be easily integrable by the methods you already know.
The methods of integration that you already know are:
1.
2.
3.
4.
5.
Integration by parts – I denote it by IP.
Ordinary Substitution – I denote it by OS.
Trigonometric Integrals – I denote it by TI.
Trigonometric Substitution – I denote it by TS.
Integration by Partial Fractions – I denote it by PF.
By a ‘proper’ function of x I mean a function in x which does not contain any trig. functions or the
natural log or the natural exponential function i.e. according to me 1+ x 2 , x 2 are all proper functions
of x even dx is a ‘proper’ function of x but e x or sin x are not ‘proper’ functions ( I shall call them
improper functions) of x even though technically they are functions of x . Note that this is not a
standard definition.
Case 1 :
 f ( x) g ( x)dx
where f(x) is a ‘proper’ function of x and g(x) is not a ‘proper’ function
of x
Approach 1 - Try IP. Try to choose u and dv in such a way such that it is easy to compute  vdu .
Remember that diffentiation decreases the power of x while integration increases the
power of x . So it is quite natural to take u=f(x) if f(x) is a power of x. On the other
hand when the ‘proper’ function is dx then it is natural to take dv = dx.
Approach 2 - Try OS.
Example :
 x sec
2
x dx
Here the proper function is x and the improper function is sec 2 x. So let us take u = x and
dv = sec 2 x dx. Then du =dx and v = tan x
We know that  udv = uv -  vdu
So
 x sec
2
x dx = x tan x -
 tan xdx
= x tan x - ln|sec x| + C
Case 2 :
 sin x cos x dx - See the book on Pg. 443
 tan x sec x dx - See the book on Pg. 445
 f ( x) g ( x)dx where f(x) = sinA or cos A and g(x) = sin B or cos B
m
m
n
n
Use the formulas given in the book on Pg. 447
Case 3 :
 f ( x)
a 2  x 2 dx or

f ( x)
a x
2
2
dx or
a
f ( x)
dx where f(x) is preferably a proper
 x2
2
function of x.
Approach 1 – Try TS . and in this case it is natural to try x=asin  so that
Case 4 :
 f ( x)
x 2  a 2 dx or

f ( x)
x2  a2
dx or
x
a 2  x 2 = acos  .
f ( x)
dx where f(x) is preferably a proper
 a2
2
function of x.
Approach 1 – Try TS . and in this case it is natural to try x=asec  so that
Case 5 :
 f ( x)
x 2  a 2 dx or

f ( x)
x2  a2
dx or
x
x 2  a 2 = atan  .
f ( x)
dx where f(x) is preferably a proper
 a2
2
function of x.
Approach 1 – Try TS . and in this case it is natural to try x=atan  so that
Example:
 5x
x 2  a 2 = asec  .
1  x 2 dx
It falls in Case 5 with a=1 . So substitute x=tan  . Therefore dx = sec 2  d  .
 5x
1  x 2 dx =
 5 tan  sec 
=
 5 tan  sec 
5  u du
=
 ax
2
dx
or
 bx  c
. u2
du
sec  tan 
2
5u 3
5 sec 3 
5(1  x 2 ) 3 / 2
+C =
+C =
+ C.
3
3
3
=
Case 6 :
sec 2  d  . Substitute u=sec  . Then du= sec  tan  d  .

dx
ax 2  bx  c
Approach 1 – For the first integral try to factor the denominator into linear factors and then try PF.
and for the second integral apply Approach 2.
Approach 2 – If the denominator is not easily factorable then try to write ax 2 +bx+c = A(x+B) 2 + C
and then apply OS of u=x+B which reduces the integral to the standard form
du
b 2
b2
2
.
With
an
easy
computation
we
can
write
ax
+bx+c
=
a(x
+
)
+
(c
)
 u2  C
2a
4a 2
b
b2
So that A = a , B =
and C = c - 2
2a
4a
Case 7 :
 dx
ax  b
dx or
 ex  f
2

ax  b
dx 2  ex  f
dx
Approach 1 – Try to write the numerator as the derivative of the denominator. Here the derivative of
dx 2 +ex + f is 2dx + e so try to get this into the numerator. One way to get it is multiply the numerator
2d
2db
2db
and the denominator by
. So the numerator changes to 2dx +
= (2dx + e) + (e)
a
a
a
a
So the first integration changes to
2d
2dx  e
a
 dx 2  ex  f dx + 2d
2db
a
 dx 2  ex  f dx .The first integral
e
can be solved by OS with u = dx 2 +ex+f while the second integral can be solved using Case 6.
Approach 2 – For the first integral try PF. But I think that the previous approach is the best approach
and solves most of the problems. For the second integral I donot think that there are any other methods
except Approach 1.
Case 8 : Refer to the Section 7.4 for all the forms of integration where you can use partial fractions.
Case 9 :
 (ax  b)
cx 2  dx  e dx
Approach 1 : This problem is similar to the second integral of Case 7 except that the quadratic
term is on the top rather than being on the bottom. As I have always mentioned , try to get the
derivative of cx 2 +dx+e in place of ax + b by algebraic manipulation. The derivative of cx 2 +dx+e is
2c
2cx + d. So multiplying and dividing the expression by
we get
a
 (ax  b)
cx 2  dx  e dx =
a
2c
 (2cx 
2cb
) cx 2  dx  e dx
a
a
2cb
(2cx  d 
 d ) cx 2  dx  e dx

a
2c
a
da
=
(2cx  d ) cx 2  dx  e dx + (b- )  cx 2  dx  e dx

2c
2c
=
The first integral can be solved by OS with u = cx 2  dx  e and the second integral can be solved by
completing the square inside the square root as shown in Case 6 Approach 2 and then use OS.
Special Case : When ax+b = 1 then just complete the square inside the square root and use OS and you
are done.
Note that in general for any kind of substitution try to get the derivative of the function that you
are substituting at the numerator of the expression that you are integrating and then try to write
all the other functions in terms of the substituted variable.
sin x cos x
For example in 
dx you could substitute u=cos x or u=sin x .In both the cases the
(1  sin x) 2
derivatives which are sin x and cos x are on the numerator but if you substitute u=cos x then you
cannot write the denominator in terms of u properly. So the best substitution should be u=sin x. Rather
the best substitution involving less computation would be u = 1+sinx
So du = cos x dx
Therefore
sin x cos x
 (1  sin x)
2
dx =

=

(u  1) cos x du
cos x
u2
u 1
du =
u2

du
u

du
1
= ln u +
+C
2
u
u
= ln |1+sin x| +
1
+ C.
1  sin x
Final tips for integration
1. Try to break up a bigger integration into several smaller integrations.
2. Use Partial Fraction only when the highest degree of the numerator is strictly less than the
highest degree of the denominator.
3. When you are substituting always try to get the derivative of the substituted variable on the
numerator of the expression.
4. Remember the standard integrations i.e. integrals 1-17 .
5. Remember the standard trigonometric identities from the front cover of the book.
6. Donot always try to substitute when you get a standard form. If possible try to reduce the
1 x2
integral to a much simpler form. For example in 
dx the natural thing to do is put
1 x2
x=sin  as there is 1-x 2 at the bottom but then the integration gets a bit complicated involving
trig functions. One could approach the integration in this way and write
2
1 x2
1 x2  2
=
(
) = -1 +
2
2
1 x2
1 x
1 x
Therefore
2
1 x2
 1  x 2 dx =   1dx +  1  x 2 dx
1
x 1
dx
| ) + C [ Since 
= -x + 2 ( ln |
is a standard integral No.19]
2
x 1
1 x2
x 1
| +C .
x 1
I did not use any substitution in this problem.
= -x + ln |
Remember that there always better and elegant methods to solve a problem depending upon the
problem , what I have written are for the general cases.