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Techniques of Integration
10.1 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 10.1.1 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 +^23 u^3 − 15 u^5 + C
= − cos x +^23 cos^3 x − 15 cos^5 x + C.
204 Chapter 10 Techniques of Integration
EXAMPLE 10.1.2 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 =^1 8
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 = − 3 2
sin 2x
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 10.1.3 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.
206 Chapter 10 Techniques of Integration
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 sin^2 x + cos^2 x = 1 in one of three forms:
cos^2 x = 1 − sin^2 x sec^2 x = 1 + tan^2 x tan^2 x = sec^2 x − 1.
If your function contains 1 − x^2 , as in the example above, try x = sin u; if it contains 1 + x^2 try x = tan u; and if it contains x^2 − 1, try x = sec u. Sometimes you will need to try something a bit different to handle constants other than one.
EXAMPLE 10.2.2 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 =^43
cos^2 u du
=^46 u + 4 sin 2 12 u+ C
= 2 arcsin(3 3 x/2)+ 2 sin^ u 3 cos^ u+ C
=
2 arcsin(3x/2) 3 +
2 sin(arcsin(3x/2)) cos(arcsin(3x/2)) 3 +^ C
= 2 arcsin(3x/2) 3
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 10.2.1.
EXAMPLE 10.2.3 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.
10.2 Trigonometric Substitutions 207
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.
10.3 Integration by Parts 209
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 10.3.1 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 10.3.2 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 10.3.3 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 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.
210 Chapter 10 Techniques of Integration
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 tan x 2 +
ln | sec x + tan x| 2 +^ C.
EXAMPLE 10.3.4 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:
sign u dv x^2 sin x − 2 x − cos x 2 − sin x − 0 cos x
or
u dv x^2 sin x − 2 x − cos x 2 − sin x 0 cos x
212 Chapter 10 Techniques of Integration
∫ 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 ⇒
10.4 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.
10.4 Rational Functions 213
EXAMPLE 10.4.1 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 10.4.2 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 10.4.3 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 −
10.4 Rational Functions 215
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 10.4.6 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.
216 Chapter 10 Techniques of Integration
Exercises 10.4.
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^ ⇒
10.5 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 10.5.1 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 10.5.1 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 10.5.2; it has area f^ (xi) + 2 f (xi+1)∆x. If we add up
218 Chapter 10 Techniques of Integration
EXAMPLE 10.5.2 Approximate
0
e−x
2 dx to two decimal places. The second deriva-
tive of f = e−x
2 is (4x^2 −2)e−x
2 , and it is not hard to see that on [0, 1], |(4x^2 −2)e−x
2 | ≤ 2. We begin by estimating the number of subintervals we are likely to need. To get two dec- imal places of accuracy, we will certainly need E(∆x) < 0 .005 or
1 12 (2)
n^2 <^0.^005 1 6 (200)^ < n
2
3 < n
With n = 6, the error estimate is thus 1/ 63 < 0 .0047. We compute the trapezoid approxi- mation for six intervals: ( f (0) 2 +^ f^ (1/6) +^ f^ (2/6) +^ · · ·^ +^ f^ (5/6) +^
f (1) 2
6 ≈^0.^74512.
So the true value of the integral is between 0. 74512 − 0 .0047 = 0.74042 and 0.74512 + 0 .0047 = 0.74982. Unfortunately, the first rounds to 0.74 and the second 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. As it turns out, we need to go to n = 12 to get two bounds that both round to the same value, which turns out to be 0.75. For comparison, using 12 rectangles to approximate the area gives 0.7727, which is considerably less accurate than the approximation using six trapezoids. In practice it generally pays to start by requiring better than the maximum possible error; for example, we might have initially required E(∆x) < 0 .001, or
1 12 (2)
n^2 <^0.^001 1 6 (1000)^ < n
2
3 < n
Had we immediately tried n = 13 this would have given us the desired answer.
The trapezoid approximation works well, especially compared to rectangles, because the tops of the trapezoids form a reasonably good approximation to the curve when ∆x is fairly small. We can extend this idea: what if we try to approximate the curve more closely,
10.5 Numerical Integration 219
by using something other than a straight line? The obvious candidate is a parabola: if we can approximate a short piece of the curve with a parabola with equation y = ax^2 + bx + c, we can easily compute the area under the parabola. There are an infinite number of parabolas through any two given points, but only one through three given points. If we find a parabola through three consecutive points (xi, f (xi)), (xi+1, f (xi+1)), (xi+2, f (xi+2)) on the curve, it should be quite close to the curve over the whole interval [xi, xi+2], as in figure 10.5.3. 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. For this to be practical, we would like a simple formula for the area under one parabola, namely, the parabola through (xi, f (xi)), (xi+1, f (xi+1)), and (xi+2, f (xi+2)). That is, we should attempt to write down the parabola y = ax^2 + bx + c through these points and then integrate it, and hope that the result is fairly simple. Although the algebra involved is messy, this turns out to be possible. 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. To find the parabola, we solve these three equations for a, b, and c:
f (xi) = a(xi+1 − ∆x)^2 + b(xi+1 − ∆x) + c f (xi+1) = a(xi+1)^2 + b(xi+1) + c f (xi+2) = a(xi+1 + ∆x)^2 + b(xi+1 + ∆x) + c
Not surprisingly, the solutions turn out to be quite messy. Nevertheless, Sage can easily compute and simplify the integral to get
∫ (^) xi+1 +∆x
xi+1 −∆x
ax^2 + bx + c dx = ∆ 3 x(f (xi) + 4f (xi+1) + f (xi+2)).
Now the sum of the areas under all parabolas is
∆x 3
(f (x 0 ) + 4f (x 1 ) + f (x 2 ) + f (x 2 ) + 4f (x 3 ) + f (x 4 ) + · · · + f (xn− 2 ) + 4f (xn− 1 ) + f (xn)) = ∆x 3 (f^ (x^0 ) + 4f^ (x^1 ) + 2f^ (x^2 ) + 4f^ (x^3 ) + 2f^ (x^4 ) +^ · · ·^ + 2f^ (xn−^2 ) + 4f^ (xn−^1 ) +^ f^ (xn)).
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. This approximation technique is referred to as Simpson’s Rule. As with the trapezoid method, this is useful only with an error estimate:
10.6 Additional exercises 221
Exercises 10.5.
In the following problems, compute the trapezoid and Simpson approximations using 4 subin- tervals, and compute the error estimate for each. (Finding the maximum values of the second and fourth derivatives can be challenging for some of these; you may use a graphing calculator or computer software to estimate the maximum values.) If you have access to Sage or similar software, approximate each integral to two decimal places. You can use this Sage worksheet to get started.
∫ (^3) 1
x dx ⇒ 2.
∫ (^3) 0
x^2 dx ⇒
∫ (^4) 2
x^3 dx ⇒ 4.
∫ (^3) 1
1 x dx^ ⇒
∫ (^2) 1
1 1 + x^2 dx^ ⇒^ 6.
∫ (^1) 0
x√1 + x dx ⇒
∫ (^5) 1
x 1 + x dx^ ⇒^ 8.
∫ (^1) 0
√ x^3 + 1 dx ⇒
∫ (^1) 0
√ x^4 + 1 dx ⇒ 10.
∫ (^4) 1
√ 1 + 1/x dx ⇒
- Using Simpson’s rule on a parabola f (x), even with just two subintervals, gives the exact value of the integral, because the parabolas used to approximate f will be f itself. Remarkably, Simpson’s rule also computes the integral of a cubic function f (x) = ax^3 + bx^2 + cx + d exactly. Show this is true by showing that ∫ x 2 x 0
f (x) dx = x^23 − · 2 x 0 (f (x 0 ) + 4f ((x 0 + x 2 )/2) + f (x 2 )). Note that the right hand side of this equation is exactly the Simpson approximation for the cubic. This does require a bit of messy algebra, so you may prefer to use Sage.
10.6 Additional exer ises
These problems require the techniques of this chapter, and are in no particular order. Some problems may be done in more than one way.
∫ (t + 4)^3 dt ⇒ 2.
∫ t(t^2 − 9)^3 /^2 dt ⇒
∫ (et^2 + 16)tet^2 dt ⇒ 4.
∫ sin t cos 2t dt ⇒
∫ tan t sec^2 t dt ⇒ 6.
∫ (^2) t + 1 t^2 + t + 3 dt^ ⇒
∫ (^1) t(t^2 − 4) dt^ ⇒^ 8.
∫ (^1) (25 − t^2 )^3 /^2 dt^ ⇒
∫ (^) cos 3t √ sin 3t dt^ ⇒^ 10.
∫ t sec^2 t dt ⇒
∫ (^) et √ et^ + 1 dt^ ⇒^ 12.
∫ cos^4 t dt ⇒
222 Chapter 10 Techniques of Integration
∫ (^1) t^2 + 3t dt^ ⇒^ 14.
∫ (^1) t^2
√ 1 + t^2 dt^ ⇒
∫ (^) sec (^2) t (1 + tan t)^3 dt^ ⇒^ 16.
∫ t^3
√ t^2 + 1 dt ⇒
∫ et^ sin t dt ⇒ 18.
∫ (t^3 /^2 + 47)^3
√ t dt ⇒
∫ (^) t 3 (2 − t^2 )^5 /^2 dt^ ⇒^ 20.
∫ (^1) t(9 + 4t^2 ) dt^ ⇒
∫ (^) arctan 2t 1 + 4t^2 dt^ ⇒^ 22.
∫ (^) t t^2 + 2t − 3 dt^ ⇒
∫ sin^3 t cos^4 t dt ⇒ 24.
∫ (^1) t^2 − 6 t + 9 dt^ ⇒
∫ (^1) t(ln t)^2 dt^ ⇒^ 26.
∫ t(ln t)^2 dt ⇒
∫ t^3 et^ dt ⇒ 28.
∫ (^) t + 1 t^2 + t − 1 dt^ ⇒
