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Techniques of Integration
Over the next few sections we examine some techniques that are frequently successful when seeking antiderivatives of functions. Sometimes this is a simple problem, since it will be apparent that the function you wish to integrate is a derivative in some straightforward way. For example, faced with (^) ∫
x^10 dx
we realize immediately that the derivative of x^11 will supply an x^10 : (x^11 )′^ = 11x^10. We don’t want the “11”, but constants are easy to alter, because differentiation “ignores” them in certain circumstances, so
d dx
11 x
11 11 x
(^10) = x (^10).
From our knowledge of derivatives, we can immediately write down a number of an- tiderivatives. Here is a list of those most often used:
∫ xn^ dx = x
n+ n + 1 +^ C,^ if^ n^6 =^ −^1 ∫ x−^1 dx = ln |x| + C ∫ ex^ dx = ex^ + C ∫ sin x dx = − cos x + C
163
164 Chapter 8 Techniques of Integration ∫ cos x dx = sin x + C ∫ sec^2 x dx = tan x + C ∫ sec x tan x dx = sec x + C ∫ (^1) 1 + x^2
dx = arctan x + C ∫ (^1) √ 1 − x^2
dx = arcsin x + C
8.1 Substitution
Needless to say, most problems we encounter will not be so simple. Here’s a slightly more complicated example: find (^) ∫
2 x cos(x^2 ) dx.
This is not a “simple” derivative, but a little thought reveals that it must have come from an application of the chain rule. Multiplied on the “outside” is 2x, which is the derivative of the “inside” function x^2. Checking:
d dx sin(x
(^2) ) = cos(x (^2) ) d dx x
(^2) = 2x cos(x (^2) ),
so (^) ∫
2 x cos(x^2 ) dx = sin(x^2 ) + C.
Even when the chain rule has “produced” a certain derivative, it is not always easy to see. Consider this problem: (^) ∫
x^3
1 − x^2 dx.
There are two factors in this expression, x^3 and
1 − x^2 , but it is not apparent that the chain rule is involved. Some clever rearrangement reveals that it is:
∫ x^3
1 − x^2 dx =
(− 2 x)
(1 − (1 − x^2 ))
1 − x^2 dx.
This looks messy, but we do now have something that looks like the result of the chain rule: the function 1 − x^2 has been substituted into −(1/2)(1 − x)
x, and the derivative
166 Chapter 8 Techniques of Integration
going on. For example, in Leibniz notation the chain rule is
dy dx =^
dy dt
dt dx.
The same is true of our current expression:
∫ (^) x 2 − 2
u du dx dx =
∫ (^) x 2 − 2
u du.
Now we’re almost there: since u = 1 − x^2 , x^2 = 1 − u and the integral is ∫ − 12 (1 − u)
u du.
It’s no coincidence that this is exactly the integral we computed in (8.1.1), we have simply renamed the variable u to make the calculations less confusing. Just as before: ∫ −
2 (1^ −^ u)
u du =
5 u^ −^
u^3 /^2 + C.
Then since u = 1 − x^2 : ∫ x^3
1 − x^2 dx =
(1 − x^2 ) − 1 3
(1 − x^2 )^3 /^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. If we can integrate this new function of u, then the antiderivative of the original function is obtained by replacing u by the equivalent expression in x. Even in simple cases you may prefer to use this mechanical procedure, since it often helps to avoid silly mistakes. For example, consider again this simple problem: ∫ 2 x cos(x^2 ) dx.
Let u = x^2 , then du/dx = 2x or du = 2x dx. Since we have exactly 2x dx in the original integral, we can replace it by du: ∫ 2 x cos(x^2 ) dx =
cos u du = sin u + C = sin(x^2 ) + C.
This is not the only way to do the algebra, and typically there are many paths to the correct answer. Another possibility, for example, is: Since du/dx = 2x, dx = du/ 2 x, and
8.1 Substitution 167
then the integral becomes ∫ 2 x cos(x^2 ) dx =
2 x cos u
du 2 x =
cos u du.
The important thing to remember is that you must eliminate all instances of the original variable x.
EXAMPLE 8.1.1 Evaluate
(ax + b)n^ dx, assuming that a and b are constants, a 6 = 0,
and n is a positive integer. We let u = ax + b so du = a dx or dx = du/a. Then ∫ (ax + b)n^ dx =
a u
n (^) du = 1 a(n + 1) u
n+1 (^) + C = 1 a(n + 1) (ax^ +^ b)
n+1 (^) + C.
EXAMPLE 8.1.2 Evaluate
sin(ax + b) dx, assuming that a and b are constants and
a 6 = 0. Again we let u = ax + b so du = a dx or dx = du/a. Then ∫ sin(ax + b) dx =
a sin^ u du^ =
a (−^ cos^ u) +^ C^ =^ −^
a cos(ax^ +^ b) +^ C.
EXAMPLE 8.1.3 Evaluate
2
x sin(x^2 ) dx. First we compute the antiderivative, then
evaluate the definite integral. Let u = x^2 so du = 2x dx or x dx = du/2. Then ∫ x sin(x^2 ) dx =
2 sin^ u du^ =
2 (−^ cos^ u) +^ C^ =^ −^
2 cos(x
2 ) + C.
Now (^) ∫ 4 2
x sin(x^2 ) dx = − 12 cos(x^2 )
4
2
= − 12 cos(16) +^12 cos(4).
A somewhat neater alternative to this method is to change the original limits to match the variable u. Since u = x^2 , when x = 2, u = 4, and when x = 4, u = 16. So we can do this: ∫ (^4)
2
x sin(x^2 ) dx =
4
2 sin^ u du^ =^ −^
2 (cos^ u)
16
4
= − 12 cos(16) +^12 cos(4).
An incorrect, and dangerous, alternative is something like this: ∫ (^4)
2
x sin(x^2 ) dx =
2
2 sin^ u du^ =^ −^
2 cos(u)
4
2
= − 12 cos(x^2 )
4
2
= − 12 cos(16) +^12 cos(4).
This is incorrect because
2
sin u du means that u takes on values between 2 and 4, which
is wrong. It is dangerous, because it is very easy to get to the point − 12 cos(u)
4
2
and forget
8.2 Powers of sine and cosine 169
8.2 Powers of sine and osine
Functions consisting of products of the sine and cosine can be integrated by using substi- tution and trigonometric identities. These can sometimes be tedious, but the technique is straightforward. Some examples will suffice to explain the approach.
EXAMPLE 8.2.1 Evaluate
sin^5 x dx. Rewrite the function:
∫ sin^5 x dx =
sin x sin^4 x dx =
sin x(sin^2 x)^2 dx =
sin x(1 − cos^2 x)^2 dx.
Now use u = cos x, du = − sin x dx:
∫ sin x(1 − cos^2 x)^2 dx =
−(1 − u^2 )^2 du
=
−(1 − 2 u^2 + u^4 ) du
= −u +^2 3
u^3 − 1 5
u^5 + C
= − cos x +^23 cos^3 x − 15 cos^5 x + C.
EXAMPLE 8.2.2 Evaluate
sin^6 x dx. Use sin^2 x = (1 − cos(2x))/2 to rewrite the
function:
∫ sin^6 x dx =
(sin^2 x)^3 dx =
∫ (^) (1 − cos 2x) 3 8 dx
=^18
1 − 3 cos 2x + 3 cos^2 2 x − cos^3 2 x dx.
Now we have four integrals to evaluate:
∫ 1 dx = x
and (^) ∫
−3 cos 2x dx = − 32 sin 2x
170 Chapter 8 Techniques of Integration
are easy. The cos^3 2 x integral is like the previous example: ∫ − cos^3 2 x dx =
− cos 2x cos^2 2 x dx
=
− cos 2x(1 − sin^2 2 x) dx
=
− 12 (1 − u^2 ) du
u − u
3 3
sin 2x − sin
(^3 2) x 3
And finally we use another trigonometric identity, cos^2 x = (1 + cos(2x))/2: ∫ 3 cos^2 2 x dx = 3
∫ (^) 1 + cos 4x 2 dx^ =
x + sin 4 4 x
So at long last we get ∫ sin^6 x dx = x 8 − 16 3 sin 2x − 161
sin 2x − sin
(^3 2) x 3
x + sin 4 4 x
+ C.
EXAMPLE 8.2.3 Evaluate
sin^2 x cos^2 x dx. Use the formulas sin^2 x = (1−cos(2x))/ 2
and cos^2 x = (1 + cos(2x))/2 to get: ∫ sin^2 x cos^2 x dx =
∫ (^1) − cos(2x) 2 ·^
1 + cos(2x) 2 dx.
The remainder is left as an exercise.
Exercises 8.2.
Find the antiderivatives.
∫ sin^2 x dx ⇒ 2.
∫ sin^3 x dx ⇒
∫ sin^4 x dx ⇒ 4.
∫ cos^2 x sin^3 x dx ⇒
∫ cos^3 x dx ⇒ 6.
∫ sin^2 x cos^2 x dx ⇒
∫ cos^3 x sin^2 x dx ⇒ 8.
∫ sin x(cos x)^3 /^2 dx ⇒
∫ sec^2 x csc^2 x dx ⇒ 10.
∫ tan^3 x sec x dx ⇒
172 Chapter 8 Techniques of Integration
EXAMPLE 8.3.2 Evaluate
4 − 9 x^2 dx. We start by rewriting this so that it looks
more like the previous example:
∫ (^) √ 4 − 9 x^2 dx =
4(1 − (3x/2)^2 ) dx =
1 − (3x/2)^2 dx.
Now let 3x/2 = sin u so (3/2) dx = cos u du or dx = (2/3) cos u du. Then
∫ 2
1 − (3x/2)^2 dx =
1 − sin^2 u (2/3) cos u du =^4 3
cos^2 u du
=^46 u + 4 sin 2 12 u+ C
= 2 arcsin(3x/2) 3
+ C
= 2 arcsin(3 3 x/2)+ 2 sin(arcsin(3x/2)) cos(arcsin(3 3 x/2))+ C
= 2 arcsin(3 3 x/2)+
2(3x/2)
1 − (3x/2)^2 3 +^ C
= 2 arcsin(3 3 x/2)+ x
4 − 9 x^2 2 +^ C,
using some of the work from example 8.3.1.
EXAMPLE 8.3.3 Evaluate
1 + x^2 dx. Let x = tan u, dx = sec^2 u du, so
∫ (^) √ 1 + x^2 dx =
1 + tan^2 u sec^2 u du =
sec^2 u sec^2 u du.
Since u = arctan(x), −π/ 2 ≤ u ≤ π/2 and sec u ≥ 0, so
sec^2 u = sec u. Then ∫ (^) √ sec^2 u sec^2 u du =
sec^3 u du.
In problems of this type, two integrals come up frequently:
sec^3 u du and
sec u du.
Both have relatively nice expressions but they are a bit tricky to discover.
8.3 Trigonometric Substitutions 173
First we do
sec u du, which we will need to compute
sec^3 u du: ∫ sec u du =
sec u sec sec^ uu^ + tan+ tan^ uu du
∫ (^) sec (^2) u + sec u tan u sec u + tan u du.
Now let w = sec u + tan u, dw = sec u tan u + sec^2 u du, exactly the numerator of the function we are integrating. Thus ∫ sec u du =
∫ (^) sec (^2) u + sec u tan u sec u + tan u du^ =
w dw^ = ln^ |w|^ +^ C = ln | sec u + tan u| + C.
Now for
sec^3 u du:
sec^3 u = sec
(^3) u 2 +
sec^3 u 2 =
sec^3 u 2 +
(tan^2 u + 1) sec u 2 = sec
(^3) u 2
(^2) u 2
= sec
(^3) u + sec u tan (^2) u 2
We already know how to integrate sec u, so we just need the first quotient. This is “simply” a matter of recognizing the product rule in action: ∫ sec^3 u + sec u tan^2 u du = sec u tan u.
So putting these together we get ∫ sec^3 u du = sec^ u 2 tan u+ ln^ |^ sec^ u 2 + tan u|+ C,
and reverting to the original variable x: ∫ (^) √ 1 + x^2 dx = sec^ u^ tan^ u 2
- ln^ |^ sec^ u^ + tan^ u| 2
+ C
= sec(arctan^ x) tan(arctan 2 x)+ ln^ |^ sec(arctan^ x) + tan(arctan 2 x)|+ C
= x
1 + x^2 2
1 + x^2 + x| 2
+ C,
using tan(arctan x) = x and sec(arctan x) =
1 + tan^2 (arctan x) =
1 + x^2.
8.4 Integration by Parts 175
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. This technique for turning one integral into another is called integration by parts, and is usually written in more compact form. If we let u = f (x) and v = g(x) then du = f ′(x) dx and dv = g′(x) dx and
∫ u dv = uv −
v du.
To use this technique we need to identify likely candidates for u = f (x) and dv = g′(x) dx.
EXAMPLE 8.4.1 Evaluate
x ln x dx. Let u = ln x so du = 1/x dx. Then we must
let dv = x dx so v = x^2 /2 and
∫ x ln x dx = x
(^2) ln x 2 −
∫ (^) x 2 2
x dx^ =^
x^2 ln x 2 −
∫ (^) x 2 dx^ =^
x^2 ln x 2 −^
x^2 4 +^ C.
EXAMPLE 8.4.2 Evaluate
x sin x dx. Let u = x so du = dx. Then we must let
dv = sin x dx so v = − cos x and ∫ x sin x dx = −x cos x −
− cos x dx = −x cos x +
cos x dx = −x cos x + sin x + C.
EXAMPLE 8.4.3 Evaluate
sec^3 x dx. Of course we already know the answer to this,
but we needed to be clever to discover it. Here we’ll use the new technique to discover the antiderivative. Let u = sec x and dv = sec^2 x dx. Then du = sec x tan x dx and v = tan x
176 Chapter 8 Techniques of Integration
and (^) ∫
sec^3 x dx = sec x tan x −
tan^2 x sec x dx
= sec x tan x −
(sec^2 x − 1) sec x dx
= sec x tan x −
sec^3 x dx +
sec x dx.
At first this looks useless—we’re right back to
sec^3 x dx. But looking more closely:
∫ sec^3 x dx = sec x tan x −
sec^3 x dx +
sec x dx ∫ sec^3 x dx +
sec^3 x dx = sec x tan x +
sec x dx
2
sec^3 x dx = sec x tan x +
sec x dx ∫ sec^3 x dx = sec^ x 2 tan x+^12
sec x dx
= sec^ x 2 tan x+ ln^ |^ sec^ x 2 + tan x|+ C.
EXAMPLE 8.4.4 Evaluate
x^2 sin x dx. Let u = x^2 , dv = sin x dx; then du = 2x dx
and v = − cos x. Now
x^2 sin x dx = −x^2 cos x +
2 x cos x dx. This is better than the
original integral, but we need to do integration by parts again. Let u = 2x, dv = cos x dx; then du = 2 and v = sin x, and
∫ x^2 sin x dx = −x^2 cos x +
2 x cos x dx
= −x^2 cos x + 2x sin x −
2 sin x dx
= −x^2 cos x + 2x sin x + 2 cos x + C.
Such repeated use of integration by parts is fairly common, but it can be a bit tedious to accomplish, and it is easy to make errors, especially sign errors involving the subtraction in the formula. There is a nice tabular method to accomplish the calculation that minimizes the chance for error and speeds up the whole process. We illustrate with the previous example. Here is the table:
178 Chapter 8 Techniques of Integration
Exercises 8.4.
Find the antiderivatives.
∫ x cos x dx ⇒ 2.
∫ x^2 cos x dx ⇒
∫ xex^ dx ⇒ 4.
∫ xex^2 dx ⇒
∫ sin^2 x dx ⇒ 6.
∫ ln x dx ⇒
∫ x arctan x dx ⇒ 8.
∫ x^3 sin x dx ⇒
∫ x^3 cos x dx ⇒ 10.
∫ x sin^2 x dx ⇒
∫ x sin x cos x dx ⇒ 12.
∫ arctan(
√ x) dx ⇒
∫ sin(√x) dx ⇒ 14.
∫ sec^2 x csc^2 x dx ⇒
8.5 Rational Fun tions
A rational function is a fraction with polynomials in the numerator and denominator. For example, x^3 x^2 + x − 6 ,^
(x − 3)^2 ,^
x^2 + 1 x^2 − 1 ,
are all rational functions of x. There is a general technique called “partial fractions” that, in principle, allows us to integrate any rational function. 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. However, in practice one does not often run across rational functions with high degree polynomials in the denominator for which one has to find the antiderivative function. So we shall explain how to find the antiderivative of a rational function only when the denominator is a quadratic polynomial ax^2 + bx + c. We should mention a special type of rational function that we already know how to integrate: If the denominator has the form (ax + b)n, the substitution u = ax + b will always work. The denominator becomes un, and each x in the numerator is replaced by (u − b)/a, and dx = du/a. While it may be tedious to complete the integration if the numerator has high degree, it is merely a matter of algebra.
8.5 Rational Functions 179
EXAMPLE 8.5.1 Find
∫ (^) x 3 (3 − 2 x)^5 dx.^ Using the substitution^ u^ = 3^ −^2 x^ we get
∫ (^) x 3 (3 − 2 x)^5 dx^ =^
( (^) u− 3 − 2
u^5 du^ =^
∫ (^) u (^3) − 9 u (^2) + 27u − 27 u^5 du
= 161
u−^2 − 9 u−^3 + 27u−^4 − 27 u−^5 du
( (^) u− 1 − 1
− 9 u
− 2 − 2
+^27 u
− 3 − 3
− 27 u
− 4 − 4
+ C
(3 − 2 x)−^1 − 1 −^
9(3 − 2 x)−^2 − 2 +
27(3 − 2 x)−^3 − 3 −^
27(3 − 2 x)−^4 − 4
+ C
= − (^) 16(3 1 − 2 x) + (^) 32(3 −^9 2 x) 2 − (^) 16(3 −^9 2 x) 3 + (^) 64(3 27 − 2 x) 4 + C
We now proceed to the case in which the denominator is a quadratic polynomial. We can always factor out the coefficient of x^2 and put it outside the integral, so we can assume that the denominator has the form x^2 + bx + c. There are three possible cases, depending on how the quadratic factors: either x^2 + bx + c = (x − r)(x − s), x^2 + bx + c = (x − r)^2 , or it doesn’t factor. We can use the quadratic formula to decide which of these we have, and to factor the quadratic if it is possible.
EXAMPLE 8.5.2 Determine whether x^2 + x + 1 factors, and factor it if possible. The quadratic formula tells us that x^2 + x + 1 = 0 when
x = −^1 ±
Since there is no square root of −3, this quadratic does not factor.
EXAMPLE 8.5.3 Determine whether x^2 − x − 1 factors, and factor it if possible. The quadratic formula tells us that x^2 − x − 1 = 0 when
x =^1 ±
Therefore
x^2 − x − 1 =
x − 1 +^
x − 1 −
8.5 Rational Functions 181
system of two equations in two unknowns. There are many ways to proceed; here’s one: If 7 = A+B then B = 7−A and so −6 = 3A− 2 B = 3A−2(7−A) = 3A−14+2A = 5A−14. This is easy to solve for A: A = 8/5, and then B = 7 − A = 7 − 8 /5 = 27/5. Thus ∫ (^7) x − 6 (x − 2)(x + 3) dx^ =
x − 2 +
x + 3 dx^ =
5 ln^ |x^ −^2 |^ +
5 ln^ |x^ + 3|^ +^ C.
The answer to the original problem is now ∫ (^) x 3 (x − 2)(x + 3) dx^ =
x − 1 dx +
∫ (^7) x − 6 (x − 2)(x + 3) dx
= x
2 2 −^ x^ +
5 ln^ |x^ −^2 |^ +
5 ln^ |x^ + 3|^ +^ C. Now suppose that x^2 + bx + c doesn’t factor. Again we can use long division to ensure that the numerator has degree less than 2, then we complete the square.
EXAMPLE 8.5.6 Evaluate
x + 1 x^2 + 4x + 8 dx.^ The quadratic denominator does not factor. We could complete the square and use a trigonometric substitution, but it is simpler to rearrange the integrand: ∫ (^) x + 1 x^2 + 4x + 8
dx =
∫ (^) x + 2 x^2 + 4x + 8
dx −
x^2 + 4x + 8
dx.
The first integral is an easy substitution problem, using u = x^2 + 4x + 8: ∫ (^) x + 2 x^2 + 4x + 8
dx =^1 2
∫ (^) du u
=^1
ln |x^2 + 4x + 8|.
For the second integral we complete the square:
x^2 + 4x + 8 = (x + 2)^2 + 4 = 4
x + 2 2
making the integral 1 4
( (^) x+ 2
dx.
Using u = x^ + 2 2 we get
1 4
( (^) x+ 2
dx =^1 4
u^2 + 1
du =^1 2
arctan
( (^) x + 2 2
The final answer is now ∫ (^) x + 1 x^2 + 4x + 8 dx^ =
2 ln^ |x
(^2) + 4x + 8| − 1 2 arctan
x + 2 2
+ C.
182 Chapter 8 Techniques of Integration
Exercises 8.5.
Find the antiderivatives.
∫ (^1) 4 − x^2 dx^ ⇒^ 2.
∫ (^) x 4 4 − x^2 dx^ ⇒
∫ (^1) x^2 + 10x + 25 dx^ ⇒^ 4.
∫ (^) x 2 4 − x^2 dx^ ⇒
∫ (^) x 4 4 + x^2 dx^ ⇒^ 6.
∫ (^1) x^2 + 10x + 29 dx^ ⇒
∫ (^) x 3 4 + x^2 dx^ ⇒^ 8.
∫ (^1) x^2 + 10x + 21 dx^ ⇒
∫ (^1) 2 x^2 − x − 3 dx^ ⇒^ 10.
∫ (^1) x^2 + 3x dx^ ⇒
8.6 Numeri al Integration
We have now seen some of the most generally useful methods for discovering antiderivatives, and there are others. Unfortunately, some functions have no simple antiderivatives; in such cases if the value of a definite integral is needed it will have to be approximated. We will see two methods that work reasonably well and yet are fairly simple; in some cases more sophisticated techniques will be needed. Of course, we already know one way to approximate an integral: if we think of the integral as computing an area, we can add up the areas of some rectangles. While this is quite simple, it is usually the case that a large number of rectangles is needed to get acceptable accuracy. A similar approach is much better: we approximate the area under a curve over a small interval as the area of a trapezoid. In figure 8.6.1 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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Figure 8.6.1 Approximating an area with rectangles and with trapezoids. (AP)
As with rectangles, we divide the interval into n equal subintervals of length ∆x. A
typical trapezoid is pictured in figure 8.6.2; it has area f^ (xi) + 2 f (xi+1)∆x. If we add up
