3.9
Lecture 14: Additional techniques of integration
This material is from Section 5.7 and Appendix G of the textbook. We have already discussed trigonometric
integrals (that is, integrals of the form
Z
sinn (x) cosm (x) dx.
• If one of n or m is odd, you use the identity sin2 (x) + cos2 (x) = 1 to change the integrand until you have just
one sin(x) or one cos(x) and the rest is the other function; then substitution for the other function works
and gives an easy polynomial.
• If n = 2 and m = 0 or n = 0 and m = 2, use trig identities: sin2 (x) =
1
2 (1 + cos(2x)).
1
2 (1
− cos(2x)) and cos2 (x) =
• If both are even and greater than zero, you can cleverly get a recursive formula using by parts, or else use a
table of integrals.
3.9.1
Trigonometric substitution
This is a special form of substitution that can work if your integrand has a term like
p
p
p
r2 − x2 , r2 + x2 , or x2 − r2 .
Example 3.92.
√ Suppose we want to measure the area of a circle of radius r. The top half of the circle is the
graph of y = r2 − x2 , so the area of the circle is equal to 4 times the area of the semi-circle, that is, of
Z rp
r2 − x2 dx.
0
This integral doesn’t lend itself to an obvious subsitution (since we don’t have an x dx) or integration by parts
(since that gives an even tougher-looking integral).
Here’s the outrageous idea which is trigonometric substitution: do the substituion θ = arcsin( xr ). Actually,
we’ll rewrite this as
x = r sin(θ), dx = r cos(θ) dθ
and when x = 0, we have θ = arcsin(0) = 0 and when x = r, we have θ = arcsin(r/r) = arcsin(1) = π/2. So the
integral becomes
Z π/2 p
r2 − r2 sin(θ) r cos(θ) dθ
0
which is nice because we have
q
p
p
2
2
r − r sin(θ) = r2 (1 − sin2 (θ)) = r2 cos2 (θ) = r cos(θ)
because on the interval [0, π/2], cos(θ) > 0 and of course r > 0 so the answer is already positive and we don’t need
an absolute value.
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Therefore, our integral is
Z
0
r
Z
p
r2 − x2 dx =
0
=
Z
π/2 p
r2 − r2 sin(θ) r cos(θ) dθ
π/2
r cos(θ) r cos(θ) dθ
0
=r
2
π/2
Z
cos2 (θ) dθ
0
= r2
Z
π/2
0
r2
=
2
Z
1
(1 + cos(2θ)) dθ
2
π/2
(1 + cos(2θ)) dθ
0
π/2
r2
1
=
θ + sin(2θ)
2
2
0
r2
[(π/2 + 0) − (0 + 0)]
2
πr2
=
4
=
whence the area of the quarter-circle is πr2 /4 of the area of a circle is πr2 , as desired.6
The trigonometric substitutions are as follows:
x = r sin(θ) and dx = r cos(θ)dθ : uses 1 − sin2 (θ) = cos2 (θ) to give
√
r2 − x2 = r cos(θ).
2
2
2
x
√ = r tan(θ) and dx = r sec (θ)dθ : uses 1 + tan (θ) = sec (θ) to give
2
2
r + x = r sec(θ).
2
2
x
√= r sec(θ) and dx = r sec(θ) tan(θ) dθ : uses sec (θ) − 1 = tan (θ) to give
x2 − r2 = r tan(θ).
(We did another example, spur of the moment, in class, which ended up being quite difficult; here is a different
one.)
Example 3.93. Evaluate
Z √
x2 − 1
dx.
x4
We run through the possibilities and settle on trigonometric substitution. The form of the term under the
square root suggests
x = sec(t) dx = sec(t) tan(t) dt
so that
q
q
p
x2 − 1 = sec( t) − 1 = tan2 (t).
It’s not true that tan(t) is always positive on the domain of sec(t); however, if we restrict to 0 < t < π/2 then
we’re fine. This corresponds to x > 1; to be safe we’d check our answer in the end to make sure it also works for
x < −1, but this is usually not a problem.
6
The ancient Greeks worked out the area of a circle with different means (see Archimedes’ “The Measurement of a Circle”) but
Archimedes worked out the surface area of a sphere using the methods of integral Calculus you’ll see in MAT1322.
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So our integral becomes
Z √
x2 − 1
dx =
x4
tan(t)
sec(t) tan(t) dt
sec4 (t)
Z
1
= tan2 (t) 3 dt
sec (t)
Z
2
sin (t)
=
cos3 (t) dt
cos2 (t)
Z
= sin2 (t) cos(t) dt
Z
= u2 du
where u = sin(t), du = cos(t)dt
Z
1
= u3 + c
3
1
= sin3 (t) + c
3
But we’re not done. We had defined x = sec(t) and need to find out what sin(t) is. Because we’re in the range
0 < t < π/2, we can just draw a right-angled triangle with base angle t representing the relationship x = sec(t).
That is,√ we set x to be the hypotenuse and 1 √
to be the adjacent side. By Pythagorus, this means the opposite
side is x2 − 1. Therefore sin(t) = opp/hyp = x2 + 1/x. Thus our final answer is
!3
√
Z √ 2
1
x −1
x2 + 1
+c
dx =
x4
3
x
which we check by differentiation.
3.9.2
Partial Fractions
See also Appendix G.
This method solves any integral in which the integrand is a rational function, that is, a polynomial divided by
a polynomial. So things that look like
Z
Z
5x2 − 4
5x − 4
dx
or
dx.
2x2 − x + 1
2x2 − x + 1
Step 1 : do long division if needed
If the degree of the numerator is greater than or equal to that of the denominator, you have to do long division
first.
Example 3.94.
15
9
5
5
1 15x − 9
5x3 − 1
4 x− 4
=
x
−
+
=
2
2
2x + x − 1
2
4 2x + x − 1
4 2x2 + x − 1
Step 2 : Factor the denominator, and complete the square in irreducible quadratic terms
If the denominator is a quadratic polynomial, then either it will factor as a product of linear factors, like
2x2 + x − 1 = (x + 1)(2x − 1)
or else it will have no real roots. You can tell that it has no real roots if you complete the square:
1
3
x2 + x + 1 = (x + )2 + .
2
4
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Since this is a sum of squares it can never be zero.
If the denominator is cubic, it has at least one root. If the root is an integer then it will divide the constant
term. For example,
0 is a root of x3 + 5x2 + 17x
and
−2 is a root of x3 + 3x2 + 3x + 2.
Given one root, you can factor your cubic into a quadratic times a linear term. For example:
x3 + 5x2 + 17x = x(x2 + 5x + 17) and x3 + 3x2 + 3x + 2 = (x + 2)(x2 + x + 1).
If the denominator has degree 4 or more, or if it has real roots which are not integers, you normally use a
computer algebra system. Obviously, in this class, we’ll restrict to things which factor nicely, or else give you the
factorization.
Step 3 : Write down the expanded form
This is the new part. Every proper (that means, the degree of the numerator is less than the degree of the
denominator) rational function can be written as a sum of simple fractions, as follows.
If your denominator factors as distinct linear factors, then the expanded form is just the inverses of each of
the factors (up to constants). So for example
4x + 3
A
B
C
=
+
+
(x − 2)(x + 2)(x − 3)
x−2 x+2 x−3
for some constants A, B, C which you need to solve for in Step 4.
If some of your factors are quadratic with no real roots, then in expanded form they have a linear term in the
numerator. For example,
2x + 3
A
Bx + C
=
+ 2
.
2
(x + 4)(x − 1)
x−1
x +4
If any factor (linear or irreducible quadratic) is repeated, then you need to include all its powers up to its
multiplicity. For example
A
B
C
Dx + E
Fx + G
2x + 3
=
+
+
+ 2
+ 2
3
2
2
2
3
(x − 1) (x + 3)
x − 1 (x − 1)
(x − 1)
x +3
(x + 3)2
Step 4 : Solve for the constants in the expanded form
The usual method for doing this is: multiply through by the common denominator on both sides and then plug
in some values for x; since the equation holds for every x, it must hold for the x you plug in, so you get some
reasonable equations.7
Example 3.95.
4x − 3
A
B
=
+
(x − 2)(x + 3)
x−2 x+3
so multiplying through by (x − 2)(x + 3) gives
4x − 3 = A(x + 3) + B(x − 2)
and if we plug in x = −3 then this gives the equation
4(−3) − 3 = 0 + B(−3 − 2) ⇒ −5B = −15 ⇒ B = 3
7
Alternatively, you can just solve the linear system of equations that you get by comparing the coefficients of xn on both sides, for
n = 0, 1, . . ..
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and if we plug in x = 2 then this gives the equation
4(2) − 3 = A(2 + 3) + 0 ⇒ 5A = 5 ⇒ A = 1
so we conclude that
1
3
4x − 3
=
+
.
(x − 2)(x + 3)
x−2 x+3
We check this answer by multiplying it out.
You see it is easy to solve for the constants when you have linear factors; when you have quadratic factors it
takes more work but you still get there.
Example 3.96.
A
Bx + C
3x2 + x + 4
=
+ 2
(x − 1)(x2 + 3)
x−1
x +3
so multiplying through gives
3x2 + x + 4 = A(x2 + 3) + (Bx + C)(x − 1).
Plugging in x = 1 gives A = 2. Plugging it x = 0 then gives
4 = 2(3) + C(−1) ⇒ C = 2
so that plugging in x = −1, for example, gives
6 = 2(4) + (−B + 2)(−2) ⇒ 6 = 8 + 2B − 4 ⇒ B = 1
and we conclude
which we check by multiplying out.
3x2 + x + 4
2
x+2
=
+ 2
2
(x − 1)(x + 3)
x−1 x +3
Step 5 : integrate
Now we can integrate any of these pieces, using substitution if needed.
Example 3.97.
Z
A
dx = A ln |x − r| + c
x−r
Example 3.98.
Z
Z
2x + 3
dx =
(x − 1)2 + 1
Z
=
Z
=
Z
=
2(x − 1) + 5
dx
(x − 1)2 + 1
2u + 5
du
u = x − 1, du = dx
u2 + 1
Z
2u
1
du
+
5
du
2
2
u +1
u +1
1
dv + 5 arctan(u)
v = u2 + 1, dv = 2udu
v
= ln((x − 1)2 + 1) + 5 arctan(x − 1) + c
as we can check by differentiation.
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= ln |v| + 5 arctan(x − 1) + c
3.9.3
A partial fractions example
These examples can be very long to do. Please see the textbook for more examples.
Example 3.99. Find
Step 1: long division. We compute
Z
2x3 + 3x2 + 22x − 6
dx
x3 + 3x2 + 12x + 10
3x2 + 2x + 26
2x3 + 3x2 + 22x − 6
=
2
−
x3 + 3x2 + 12x + 10
x3 + 3x2 + 12x + 10
Step 2: factor. We see that −1 is a root of the denominator, so we divide through by x + 1 to get
x3 + 3x2 + 12x + 10 = (x + 1)(x2 + 2x + 10) = (x + 1)((x + 1)2 + 9).
Step 3: write expanded formof proper fraction. We have
A
Bx + C
3x2 + 2x + 26
=
+
x3 + 3x2 + 12x + 10
x + 1 (x + 1)2 + 9
Step 4: solve for coefficients. We cross-multiply to get
3x2 + 2x + 26 = A((x + 1)2 + 9) + (Bx + C)(x + 1).
Plug in x = −1 to get 3 − 2 + 26 = 9A or A = 3. Plug in x = 0 to get 26 = 3(10) + C(1) or C = −4. Plug in
x = 1, for example, to get 31 = 3(13) + (B − 4)(2) or B = 0. Thus we have
3x2 + 2x + 26
3
−4
=
+
3
2
x + 3x + 12x + 10
x + 1 (x + 1)2 + 9
Step 5: integrate. We have
Z Z
3x2 + 2x + 26
2x3 + 3x2 + 22x − 6
2
−
dx
=
dx
x3 + 3x2 + 12x + 10
x3 + 3x2 + 12x + 10
Z −4
3
+
dx
= 2x −
x + 1 (x + 1)2 + 9
Z
Z
3
1
= 2x −
dx + 4
dx
x+1
(x + 1)2 + 9
Z
1
= 2x − 3 ln |x + 1| + 4
du u = x + 1, du = dx.
u2 + 9
This last integral can be solved by noting that
1
1
1
1
=
=
u2 + 9
9((u/3)2 + 1)
9 (u/3)2 + 1
so that v = u/3, dv = 31 du or du = 3dv yields
Z
Z
1
1
1
3
1
du =
dv = arctan(v) + c = arctan((x + 1)/3) + c
2
2
u +9
9 v +1
3
3
so that our final answer is
2x − 3 ln |x + 1| +
4
arctan((x + 1)/3) + c
3
as we can check by differentiation.
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