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8
Techniques of Integration

Over the next few sections we examine some techniques that are frequently successful when
seeking antiderivatives of functions
...
For example, faced with
x10 dx
we realize immediately that the derivative of x11 will supply an x10 : (x11 )′ = 11x10
...

dx 11
11
From our knowledge of derivatives, we can immediately write down a number of antiderivatives
...
Here’s a slightly more
complicated example: find
2x cos(x2 ) dx
...
Multiplied on the “outside” is 2x, which is the derivative
of the “inside” function x2
...

Even when the chain rule has “produced” a certain derivative, it is not always easy to
see
...


There are two factors in this expression, x3 and 1 − x2 , but it is not apparent that the
chain rule is involved
...


This looks messy, but we do now have something that looks like the result of the chain
√
rule: the function 1 − x2 has been substituted into −(1/2)(1 − x) x, and the derivative

8
...
If we can find a function F (x) whose derivative
√
is −(1/2)(1 − x) x we’ll be done, since then
1
d
F (1 − x2 ) = −2xF ′ (1 − x2 ) = (−2x) −
dx
2
= x3

(1 − (1 − x2 )) 1 − x2

1 − x2

But this isn’t hard:
√
1
− (1 − x) x dx =
2

1
− (x1/2 − x3/2 ) dx
2
2 3/2 2 5/2
x − x
3
5

=−

1
2

=

1
1
x−
5
3

(8
...
1)
+C

x3/2 + C
...


So we succeeded, but it required a clever first step, rewriting the original function so
that it looked like the result of using the chain rule
...
It does sometimes not work, or may require more than one attempt, but the
idea is simple: guess at the most likely candidate for the “inside function”, then do some
algebra to see what this requires the rest of the function to look like
...
Now we know that the chain rule will multiply by the derivative of this inner
function:
du
= −2x,
dx
so we need to rewrite the original function to include this:
x3

1 − x2 =

√ −2x
x3 u
dx =
−2x

x2 √ du
dx
...
For example, in Leibniz notation the chain rule is
dy dt
dy
=

...

−2

Now we’re almost there: since u = 1 − x2 , x2 = 1 − u and the integral is
√
1
− (1 − u) u du
...
1
...
Just as before:
√
1
− (1 − u) u du =
2

1
1
u−
5
3

u3/2 + C
...


To summarize: if we suspect that a given function is the derivative of another via the
chain rule, we let u denote a likely candidate for the inner function, then translate the
given function so that it is written entirely in terms of u, with no x remaining in the
expression
...

Even in simple cases you may prefer to use this mechanical procedure, since it often
helps to avoid silly mistakes
...

Let u = x2 , then du/dx = 2x or du = 2x dx
...


This is not the only way to do the algebra, and typically there are many paths to the
correct answer
...
1

Substitution

167

then the integral becomes
2x cos(x2 ) dx =

2x cos u

du
=
2x

cos u du
...

(ax + b)n dx, assuming that a and b are constants, a = 0,

Evaluate

EXAMPLE 8
...
1

and n is a positive integer
...
Then
(ax + b)n dx =

1 n
1
1
u du =
un+1 + C =
(ax + b)n+1 + C
...
1
...
Again we let u = ax + b so du = a dx or dx = du/a
...

a
a
a

sin(ax + b) dx =

4

x sin(x2 ) dx
...
1
...
Let u = x2 so du = 2x dx or x dx = du/2
...

2
2
2

x sin(x2 ) dx =
Now
4

1
x sin(x ) dx = − cos(x2 )
2

4

2

2

2

1
1
= − cos(16) + cos(4)
...
Since u = x2 , when x = 2, u = 4, and when x = 4, u = 16
...

2
2

An incorrect, and dangerous, alternative is something like this:
4

4

2

x sin(x ) dx =
2

2

1
1
sin u du = − cos(u)
2
2
4

This is incorrect because
2

4
2

1
= − cos(x2 )
2

4
2

1
1
= − cos(16) + cos(4)
...
It is dangerous, because it is very easy to get to the point − cos(u)
2

4

and forget
2

168

Chapter 8 Techniques of Integration

1
1
to substitute x2 back in for u, thus getting the incorrect answer − cos(4) + cos(2)
...

2
2

1/2

cos(πt)
dt
...
We change the limits to sin(π/4) = 2/2 and sin(π/2) = 1
...
1
...

=− +
π
π

Exercises 8
...

Find the antiderivatives or evaluate the definite integral in each problem
...


(1 − t)9 dt ⇒

2
...


x(x2 + 1)100 dx ⇒

4
...


sin3 x cos x dx ⇒

6
...


cos(πt) cos sin(πt) dt ⇒

x2
√
dx ⇒
1 − x3
sin x
dx ⇒
cos3 x

7
...


π

11
...

0
4

15
...


sin5 (3x) cos(3x) dx ⇒
π/2

x sec2 (x2 ) tan(x2 ) dx ⇒

1
dx ⇒
(3x − 7)2

6x
dx ⇒
(x2 − 7)1/9

−1

sin7 x dx ⇒

1
dt ⇒
1 − 5t
100 − x2 dx ⇒

10
...


sec2 x tan x dx ⇒

14
...

0

(cos2 x − sin2 x) dx ⇒

1

18
...


(x2 + 1)2 dx ⇒

20
...
2

º¾

ÈÓÛ Ö× Ó

× Ò

Ò

Powers of sine and cosine

169

Ó× Ò

Functions consisting of products of the sine and cosine can be integrated by using substitution and trigonometric identities
...
Some examples will suffice to explain the approach
...
2
...
Rewrite the function:

sin x sin4 x dx =

sin x(sin2 x)2 dx =

sin x(1 − cos2 x)2 dx
...

3
5

EXAMPLE 8
...
2

Evaluate

sin6 x dx
...


Now we have four integrals to evaluate:
1 dx = x
and

3
−3 cos 2x dx = − sin 2x
2

170

Chapter 8 Techniques of Integration

are easy
...


And finally we use another trigonometric identity, cos2 x = (1 + cos(2x))/2:
3 cos2 2x dx = 3

1 + cos 4x
3
dx =
2
2

x+

sin 4x
4


...
2
...


sin2 x cos2 x dx
...

2
2

The remainder is left as an exercise
...
2
...

1
...


sin3 x dx ⇒

3
...


cos2 x sin3 x dx ⇒

5
...


sin2 x cos2 x dx ⇒

7
...


sin x(cos x)3/2 dx ⇒

9
...


tan3 x sec x dx ⇒

8
...
Occasionally it can help to replace the original variable by something
more complicated
...

EXAMPLE 8
...
1

Evaluate

1 − x2 dx =

1 − x2 dx
...
Then
1 − sin2 u cos u du =

√

cos2 u cos u du
...
Consider again the substitution x = sin u
...
If we do, then by the definition of the arcsine, −π/2 ≤ u ≤ π/2, so
cos u ≥ 0
...

2
4

This is a perfectly good answer, though the term sin(2 arcsin x) is a bit unpleasant
...
Using the identity sin 2x = 2 sin x cos x, we can write sin 2u =
2 sin u cos u = 2 sin(arcsin x)
full antiderivative is

1 − sin2 u = 2x 1 − sin2 (arcsin x) = 2x 1 − x2
...

2
4
2
2
This type of substitution is usually indicated when the function you wish to integrate
contains a polynomial expression that might allow you to use the fundamental identity
sin2 x + cos2 x = 1 in one of three forms:
cos2 x = 1 − sin2 x

sec2 x = 1 + tan2 x

tan2 x = sec2 x − 1
...
Sometimes you will need to try
something a bit different to handle constants other than one
...
3
...
We start by rewriting this so that it looks

more like the previous example:
4 − 9x2 dx =

4(1 − (3x/2)2 ) dx =

1 − (3x/2)2 dx
...
Then
2

1 − (3x/2)2 dx =

2

1 − sin2 u (2/3) cos u du =

4
3

cos2 u du

4u 4 sin 2u
+
+C
6
12
2 arcsin(3x/2) 2 sin u cos u
=
+
+C
3
3
2 arcsin(3x/2) 2 sin(arcsin(3x/2)) cos(arcsin(3x/2))
=
+
+C
3
3
=

2 arcsin(3x/2) 2(3x/2) 1 − (3x/2)2
+
+C
3
3
√
2 arcsin(3x/2) x 4 − 9x2
=
+
+ C,
3
2
=

using some of the work from example 8
...
1
...
3
...
Let x = tan u, dx = sec2 u du, so

1 + tan2 u sec2 u du =

1 + x2 dx =

Since u = arctan(x), −π/2 ≤ u ≤ π/2 and sec u ≥ 0, so
√

sec2 u sec2 u du =

√

√

sec2 u sec2 u du
...
Then

sec3 u du
...


sec u du
...
3

First we do

Trigonometric Substitutions

sec u du, which we will need to compute
sec u du =
=

sec u

173

sec3 u du:

sec u + tan u
du
sec u + tan u

sec2 u + sec u tan u
du
...
Thus
sec u du =

sec2 u + sec u tan u
du =
sec u + tan u

1
dw = ln |w| + C
w

= ln | sec u + tan u| + C
...

2
2
2
2
2
3

We already know how to integrate sec u, so we just need the first quotient
...

So putting these together we get
sec3 u du =

sec u tan u ln | sec u + tan u|
+
+ C,
2
2

and reverting to the original variable x:
1 + x2 dx =

sec u tan u ln | sec u + tan u|
+
+C
2
2

sec(arctan x) tan(arctan x) ln | sec(arctan x) + tan(arctan x)|
+
+C
2
2
√
√
ln | 1 + x2 + x|
x 1 + x2
+
+ C,
=
2
2
=

using tan(arctan x) = x and sec(arctan x) =

1 + tan2 (arctan x) =

1 + x2
...
3
...

1
...

5
...

9
...


º

2
...


1 − x2 dx ⇒

6
...

10
...


Ö Ø ÓÒ

1 − x2 dx ⇒
x2 + 2x dx ⇒

x2
√
dx ⇒
4 − x2
x3
√
dx ⇒
4x2 − 1

Ý È ÖØ×

We have already seen that recognizing the product rule can be useful, when we noticed
that
sec3 u + sec u tan2 u du = sec u tan u
...

Start with the product rule:
d
f (x)g(x) = f ′ (x)g(x) + f (x)g ′ (x)
...


f (x)g(x) =
and then

This may not seem particularly useful at first glance, but it turns out that in many cases
we have an integral of the form
f (x)g ′ (x) dx
but that
f ′ (x)g(x) dx
is easier
...
If we let u = f (x) and v = g(x) then

8
...


To use this technique we need to identify likely candidates for u = f (x) and dv = g ′ (x) dx
...
4
...
Let u = ln x so du = 1/x dx
...
4
...

2
2
4

x sin x dx
...
Then we must let

dv = sin x dx so v = − cos x and
x sin x dx = −x cos x −

EXAMPLE 8
...
3

− cos x dx = −x cos x +

Evaluate

cos x dx = −x cos x + sin x + C
...
Of course we already know the answer to this,

but we needed to be clever to discover it
...
Let u = sec x and dv = sec2 x dx
...


176

Chapter 8 Techniques of Integration

sec3 x dx
...

2
2

x2 sin x dx = −x2 cos x +

and v = − cos x
...
Let u = x2 , dv = sin x dx; then du = 2x dx

Evaluate

EXAMPLE 8
...
4

sec x tan x 1
+
2
2

sec x dx

2x cos x dx
...
Let u = 2x, dv = cos x dx;
then du = 2 and v = sin x, and
x2 sin x dx = −x2 cos x +

2x cos x dx

= −x2 cos x + 2x sin x −

2 sin x dx

= −x2 cos x + 2x sin x + 2 cos x + C
...
There is a nice tabular method to accomplish the calculation that minimizes
the chance for error and speeds up the whole process
...
Here is the table:
sign

u

dv

u

dv

x2

sin x

x2

sin x

−

2x

− cos x

−2x

− cos x

−

0

0

cos x

2

− sin x
cos x

or

2

− sin x

8
...
In the first column, we place a “−” in every second row
...

To compute with this second table we begin at the top
...
This gives:
−x2 cos x +

2x cos x dx,

or exactly the result of the first application of integration by parts
...
Now we multiply twice on the diagonal, (x2 )(− cos x)
and (−2x)(− sin x) and then once straight across, (2)(− sin x), and combine these as
−x2 cos x + 2x sin x −

2 sin x dx,

giving the same result as the second application of integration by parts
...
Now multiply three times on the diagonal
to get (x2 )(− cos x), (−2x)(− sin x), and (2)(cos x), and once straight across, (0)(cos x)
...


Typically we would fill in the table one line at a time, until the “straight across” multiplication gives an easy integral
...


Exercises 8
...

Find the antiderivatives
...


x cos x dx ⇒

2
...


xex dx ⇒

4
...


sin2 x dx ⇒

6
...


x arctan x dx ⇒

9
...


x3 sin x dx ⇒

x3 cos x dx ⇒

10
...


x sin x cos x dx ⇒

12
...


√
sin( x) dx ⇒

14
...

For example,
x3
1
x2 + 1
,
,
,
x2 + x − 6
(x − 3)2
x2 − 1
are all rational functions of x
...
The algebraic steps in the
technique are rather cumbersome if the polynomial in the denominator has degree more
than 2, and the technique requires that we factor the denominator, something that is not
always possible
...
So we shall explain how to find the antiderivative of a rational function only
when the denominator is a quadratic polynomial ax2 + bx + c
...
The denominator becomes un , and each x in the numerator is replaced by
(u − b)/a, and dx = du/a
...


8
...
Using the substitution u = 3 − 2x we get
(3 − 2x)5

Find

EXAMPLE 8
...
1

179

Rational Functions

u−3
−2

u5

3

du =

1
16

u3 − 9u2 + 27u − 27
du
u5

u−2 − 9u−3 + 27u−4 − 27u−5 du
u−1
9u−2
27u−3
27u−4
−
+
−
−1
−2
−3
−4

+C

(3 − 2x)−1
9(3 − 2x)−2
27(3 − 2x)−3
27(3 − 2x)−4
−
+
−
−1
−2
−3
−4

+C

1
9
9
27
+
−
+
+C
2
3
16(3 − 2x) 32(3 − 2x)
16(3 − 2x)
64(3 − 2x)4

We now proceed to the case in which the denominator is a quadratic polynomial
...
There are three possible cases, depending
on how the quadratic factors: either x2 + bx + c = (x − r)(x − s), x2 + bx + c = (x − r)2 ,
or it doesn’t factor
...

EXAMPLE 8
...
2 Determine whether x2 + x + 1 factors, and factor it if possible
...


Since there is no square root of −3, this quadratic does not factor
...
5
...
The
quadratic formula tells us that x2 − x − 1 = 0 when
x=
Therefore
x2 − x − 1 =

1±

√
1+4
1± 5
=

...


180

Chapter 8 Techniques of Integration

If x2 + bx + c = (x − r)2 then we have the special case we have already seen, that can
be handled with a substitution
...

If x2 + bx + c = (x − r)(x − s), we have an integral of the form
p(x)
dx
(x − r)(x − s)
where p(x) is a polynomial
...

x3
dx in terms of an integral with a numerator
(x − 2)(x + 3)
that has degree less than 2
...
5
...

(x − 2)(x + 3)

The first integral is easy, so only the second requires some work
...

x−r x−s
(x − r)(x − s)
(x − r)(x − s)
That is, adding two fractions with constant numerator and denominators (x−r) and (x−s)
produces a fraction with denominator (x − r)(x − s) and a polynomial of degree less than
2 for the numerator
...
An example should make it clear how
to proceed
...
5
...
We start by writing
(x − 2)(x + 3)
(x − 2)(x + 3)
as the sum of two fractions
...

(x − 2)(x + 3)
x−2 x+3
If we go ahead and add the fractions on the right hand side we get
(A + B)x + 3A − 2B
7x − 6
=

...
This is a problem you’ve seen before: solve a

8
...
There are many ways to proceed; here’s one: If
7 = A+B then B = 7−A and so −6 = 3A−2B = 3A−2(7−A) = 3A−14+2A = 5A−14
...
Thus
7x − 6
dx =
(x − 2)(x + 3)

27 1
8
27
8 1
+
dx = ln |x − 2| +
ln |x + 3| + C
...

2
5
5

Now suppose that x2 + bx + c doesn’t factor
...

x+1
dx
...
We could complete the square and use a trigonometric substitution, but it is simpler
to rearrange the integrand:
x+2
1
x+1
dx =
dx −
dx
...
5
...

2 + 4x + 8
x
2
u
2
For the second integral we complete the square:
x+2
2

x2 + 4x + 8 = (x + 2)2 + 4 = 4

2

+1 ,

making the integral
1
4
Using u =

x+2
we get
2
1
1
1
dx =
2
x+2
4
4
+1
2

1
x+2 2
2

u2

+1

dx
...


+ C
...
5
...

1
dx ⇒
1
...

dx ⇒
x2 + 10x + 25
x4
5
...

dx ⇒
4 + x2
1
dx ⇒
9
...

4
...

8
...


x4
dx ⇒
4 − x2
x2
dx ⇒
4 − x2
1
dx ⇒
2 + 10x + 29
x
1
dx ⇒
x2 + 10x + 21
1
dx ⇒
2 + 3x
x

Ö Ø ÓÒ

We have now seen some of the most generally useful methods for discovering antiderivatives,
and there are others
...
We will
see two methods that work reasonably well and yet are fairly simple; in some cases more
sophisticated techniques will be needed
...
While this
is quite simple, it is usually the case that a large number of rectangles is needed to get
acceptable accuracy
...
In figure 8
...
1 we see an area under
a curve approximated by rectangles and by trapezoids; it is apparent that the trapezoids
give a substantially better approximation on each subinterval
...


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6
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As with rectangles, we divide the interval into n equal subintervals of length ∆x
...
6
...
If we add up
2

8
...

+ f (x1 ) + f (x2 ) + · · · + f (xn−1 ) +
2
2
This is usually known as the Trapezoid Rule
...


...


...



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...


...


...


...


...


...


...


...
6
...


In practice, an approximation is useful only if we know how accurate it is; for example,
we might need a particular value accurate to three decimal places
...
For any approximation technique, we need an
error estimate, a value that is guaranteed to be larger than the actual error
...
In the case of our approximation of the
integral, we want E = E(∆x) to be a function of ∆x that gets small rapidly as ∆x gets
small
...

THEOREM 8
...
1 Suppose f has a second derivative f ′′ everywhere on the interval
[a, b], and |f ′′ (x)| ≤ M for all x in the interval
...


b−a
(b − a)3
M (∆x)2 =
M
...
6
...
The second deriva-

Approximate
0

−x2

2

−x2

2

tive of f = e
is (4x −2)e
, and it is not hard to see that on [0, 1], |(4x2 −2)e−x | ≤ 2
...
To get two decimal places of accuracy, we will certainly need E(∆x) < 0
...
005
12
n
1
(200) < n2
6
5
...
0047
...
74512
...
74512 − 0
...
74042 and 0
...
0047 = 0
...
Unfortunately, the first rounds to 0
...
75,
so we can’t be sure of the correct value in the second decimal place; we need to pick a larger
n
...
75
...
7727, which is considerably less accurate than the approximation using six
trapezoids
...
001, or
1
1
(2) 2 < 0
...
91 ≈

500
3

Had we immediately tried n = 13 this would have given us the desired answer
...
We can extend this idea: what if we try to approximate the curve more closely,

8
...

There are an infinite number of parabolas through any two given points, but only
one through three given points
...
6
...
If we divide the interval [a, b]
into an even number of subintervals, we can then approximate the curve by a sequence of
parabolas, each covering two of the subintervals
...
That is, we should attempt to write down the parabola
y = ax2 + bx + c through these points and then integrate it, and hope that the result is
fairly simple
...
The
algebra is well within the capability of a good computer algebra system like Sage, so we
will present the result without all of the algebra; you can see how to do it in this Sage
worksheet
...
Nevertheless, Sage can easily
compute and simplify the integral to get
xi+1 +∆x

ax2 + bx + c dx =
xi+1 −∆x

∆x
(f (xi ) + 4f (xi+1 ) + f (xi+2 ))
...

3
This is just slightly more complicated than the formula for trapezoids; we need to remember
the alternating 2 and 4 coefficients; note that n must be even for this to make sense
...

As with the trapezoid method, this is useful only with an error estimate:

186

Chapter 8 Techniques of Integration

(xi , f (xi ))


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6
...


THEOREM 8
...
3 Suppose f has a fourth derivative f (4) everywhere on the interval
[a, b], and |f (4) (x)| ≤ M for all x in the interval
...

180
180n4

1

EXAMPLE 8
...
4

2

e−x dx to two decimal places
...
We begin by estimating the number of subintervals we are likely to need
...
005, but taking
a cue from our earlier example, let’s require E(∆x) < 0
...
001
180
n
200
< n4
3
2
...
Then the error estimate
is 12/180/44 < 0
...
746855
...
746855 − 0
...
746555 and 0
...
0003 = 0
...
75
...
7

Additional exercises

187

Exercises 8
...

In the following problems, compute the trapezoid and Simpson approximations using 4 subintervals, and compute the error estimate for each
...
) If you have access to Sage or similar
software, approximate each integral to two decimal places
...

3

3

x dx ⇒

1

4
...


x
dx ⇒
1+x

8
...

2
2

5
...

1

x2 dx ⇒

2
...


3

1
dx ⇒
x
1
1 √
x 1 + x dx ⇒

0

1

x3 + 1 dx ⇒

0
4

1

x4 + 1 dx ⇒

9
...

1

1 + 1/x dx ⇒

11
...
Remarkably,
Simpson’s rule also computes the integral of a cubic function f (x) = ax3 + bx2 + cx + d
exactly
...

f (x) dx =
3·2
x0
This does require a bit of messy algebra, so you may prefer to use Sage
...
Some
problems may be done in more than one way
...


(t + 4)3 dt ⇒

3
...

7
...

11
...


t(t2 − 9)3/2 dt ⇒

(et + 16)tet dt ⇒

4
...


2

2

1
dt ⇒
t(t2 − 4)
cos 3t
√
dt ⇒
sin 3t
et
√
dt ⇒
et + 1

8
...


t sec2 t dt ⇒

12
...


Chapter 8 Techniques of Integration

t2

1
dt ⇒
+ 3t

14
...


sec t
dt ⇒
(1 + tan t)3

16
...


et sin t dt ⇒

18
...

21
...

22
...


sin3 t cos4 t dt ⇒

24
...


1
dt ⇒
t(ln t)2

26
...


t3 et dt ⇒

28

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Title: Integral of a function
Description: Here you will learn all techniques of Integration. And i am sure you will be able to integrate all the functions you work with . Hope you like it !! .

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