Replace tan2xby sec2x− 1 and split the integral into the difference of two integrals, sec3x dx− sec x dx.. The method of partial fractions is an algebraic technique enabling us to split
Trang 129.2 Trigonometric Integrals and Trigonometric Substitution 891 1
√
9− x2=
1 3 9
9−x2
9
=1 3 1
1−x
3
2
1
√
9− x2dx=
1
1−x
3
213dx Let u=x
3, du=1
3 dx,
= arcsin
x
3
+ C
Method 2: We’d love to get rid of that square root in the denominator To do this, let’s replace 9− x2by a perfect square We can exploit the trigonometric identity
1− sin2θ= cos2θ or 9− 9 sin2θ= 9 cos2θ
Let x = 3 sin θ, where θ ∈
−π
2,
π
2
, dx = 3 cos θdθ.
Then√
9− x2=9− 9 sin2θ=√9 cos2θ= 3√cos2θ
1
√
9− x2dx=
1
3√ cos2θ 3 cos θ dθ1
=
1 dθ
= θ + C But x = 3 sin θ, so x
3= sin θ and θ = arcsinx
3
1
√
9− x2dx= arcsin
x
3
+ C
This second method will come in very handy, particularly when the first method cannot
be applied Notice that although the integrand involves no trigonometry, the antiderivative does ◆
0
√
4− x2dx
SOLUTION The strategy is to replace 4− x2by a perfect square in order to eliminate the square root
Let x = 2 sin θ, where θ ∈
−π
2 ,
π
2
, dx = 2 cos θ dθ.
4− x2= 4− 4 sin2θ=4 cos2θ= 2cos2θ
What are the new endpoints of integration?
When x= 0, 0= 2 sin θ ⇒ sin θ = 0 ⇒ θ = arcsin 0 = 0.
When x= 1, 1= 2 sin θ ⇒ sin θ =1
2 ⇒ θ = arcsin1
2=π
6
1 √ cos 2θis actually| cos θ| as opposed to cos θ In the context of this problem, however,√cos 2θ can be replaced by cos θ because θ has been restricted to
−π,π
and cos θ≥ 0 on this interval.
Trang 21 0
4− x2dx=
6
0
2 cos2θ 2 cos θ dθ
= 4
π
6
0
cos2θ dθ Use cos2θ=1
2[1+ cos 2θ].
= 2
π
6
0
(1+ cos 2θ) dθ
= 2
θ+1
2 sin 2θ
π
6
0
= 2
π
6 +1
2 sin
π
3 − 0
=π
3 +1 2
√ 3 2
Clarifications and Justifications
Notice that when we let x = 2 sin θ, we are restricting x to lie between −2 and 2 That’s
fine, because√
4− x2only makes sense for x in this interval.
In both of the previous examples we’ve said √
cos2θ = cos θ Actually,√cos2θ=
| cos θ|, which is equal to cos θ only if cos θ is nonnegative It is because we’ve restricted
θto [−π
2
π
2] that we can write| cos θ| = cos θ.
In fact, the substitution x = g(θ) is only valid if g(θ) is 1-to-1 In other words, when we say x = 2 sin θ we must restrict θ In this case we choose θ ∈−π
2 ,π2 , the same restrictions used for sin−1x This type of restriction will hold in all the examples that follow
Notice that once we’ve established that θ∈−π
2,π2
we can say, as above, if sin θ= 0,
then θ= 0 ◆
Generalizing the Method of Trigonometric Substitution: k is a constant.
k2− u2 u = k sin θ k2− u2= C cos θ θ∈−π
2,π2
k2+ u2 u = k tan θ k2+ u2= C sec θ θ∈−π
2,π2
u2− k2 u = k sec θ u2− k2= C tan θ θ∈0,π2
,
π,3π2
The restriction on θ for the substitution u = k sec θ is designed to guarantee that
| tan θ| = tan θ.2Restrictions on θ specified above will hold in the examples that follow.
CAUTION As always, try to make your work as simple as possible The table supplied above
does not mean you ought to use trigonometric substitution on x
√
x2 −4dx, for instance.
Use ordinary substitution, u = x2− 4, instead
2
Trang 329.2 Trigonometric Integrals and Trigonometric Substitution 893
x2√
9+x2 dx
2,π2
9+ x2=9+ 9 tan2θ= 31+ tan2θ= 3sec2θ = 3 sec θ
1
x2√
9+ x2dx=
1
9 tan2θ 3 sec θ · 3 sec2θ dθ
=1 9
sec θ
tan2θ dθ
=1 9
cos2θ cos θ sin2θ dθ
=1 9
cos θ
sin2θ dθ Let u = sin θ, du = cos θ dθ.
=1 9
1
u2du
=1 9
u−2du
= −1
9u
−1+ C
9 sin θ + C Now we must change from θ back to x It is not appealing to write sin θ= sintan−1 x
3
without simplifying Instead, we can use a triangle to sort out the conversion to x as follows:
x = 3 tan θ ⇒ tan θ = x
3 Draw a right triangle, label θ , and label the sides so that tan θ=x
3
We determine the hypotenuse using the Pythagorean Theorem (See Figure 29.1)
3
θ
3
θ
9 + x2
Figure 29.1
sin θ=√ x
9+ x2
Trang 4x2√
9+ x2dx= −1
9
1
sin θ + C
= −1 9
√
9+ x2
= −
√
9+ x2
9x + C
Check this by differentiating ◆
4− x2dx
x = 2 sin θ and find
4− x2dx= 2
θ+1
2 sin 2θ
+ C
= 2θ + sin 2θ + C.
The task now is to change from θ back to x We’d like to use a triangle, as in the previous example, but we must first apply the trigonometric identity sin 2θ = 2 sin θ cos θ.
4− x2dx = 2θ + 2 sin θ cos θ + C
x = 2 sin θ ⇒ sin θ = x
2 Draw a right triangle, label θ , and label the sides so that sin θ=x
2 (See Figure 29.2.)
4− x2dx = 2θ + 2 sin θ cos θ + C
= 2 arcsin
x
2
+ 2
x
2
√
4− x2 2
+ C
= 2 arcsin
x
2
+x
√
4− x2
⇒
√4 – x2
Figure 29.2 ◆ Can problems become more involved than this? Of course they can Consider, for instance, √
4x − x2dx.√
4x − x2can be expressed as
4− (x − 2)2by completing the square.
4− (x − 2)2dx is of the form √
4− u2duand we continue as in the previous example
At some point, when the level of complexity overpowers the level of intrigue, you’ll turn to a computer or powerful calculator Regardless, it is useful to have this method of trigonometric substitution in your array of tools
Trang 529.2 Trigonometric Integrals and Trigonometric Substitution 895
P R O B L E M S F O R S E C T I O N 2 9 2
In Problems 1 through 18, evaluate the integral.
1.
cos2x dx
2.π
0 sin3x dx
3.
cos x sin2x dx
4.
cos3xsin2x dx
5.
cos4x dx
6.
cos2xsin2x dx
7.
cos4xsin3x dx
8.
cos3xsin11x dx
9.
cos3( 3x) dx
10.π
2
0 cos5x√
sin x dx
11. sin x
√
cos 3x dx
12.
tan 3x dx
13.
tan 2x sec 2x dx
14.π
4
0
√
tan x sec2x dx
15.
tan x sec4x dx
16.
tan3xsec4x dx
17.
tan3xsec5x dx
18.
tan3x sec x dx
19. tan3x
sec 4x dx
20.
tan8xsec4x dx
21. sin x
cos 2x dx
Problems 22 through 24 refer to the information provided about Fourier series A Fourier series expresses a function as a weighted infinite sum of terms of the form
Trang 6sin nx and cos nx, where n is a nonnegative integer Fourier series are a very powerful tool In order to construct such a series we need the following results m and n are positive integers.
(a) π
−π sin mx cos nx dx= 0
(b) π
−π sin mx sin nx dx=
π if m = n
0 if m = n (c)π
−π cos mx cos nx dx=
π if m = n
0 if m = n These results can be obtained using the product formulas given in this section.
22 Prove statement (a) above
23 Prove statement (b) above
24 Prove statement (c) above
25 Find the volume generated by revolving the region under one arch of cos x about the
x-axis
26 Find the volume generated by revolving the region between the graphs of y = sin x and
y = cos x from x = π
4 to x=5π
4 around the horizontal line y= 2
27 Find 1
sin θ dθ
28 Evaluate
sec3x dxusing a reduction formula
In Problems 29 through 43, evaluate the integrals Not all require a trigonometric substitution Choose the simplest method of integration.
29.
x3√
4− x2dx
30. x3
√
9−x2 dx
31. x
√
4+x2 dx
32. x3
√
4+x2 dx
33. 1
√
9+x2 dx
34.
x2√
x2− 9 dx
35.3
0
√
9− 4x2dx
36. √
4− 9x2dx
37.
x√
4− 9x2dx
Trang 729.2 Trigonometric Integrals and Trigonometric Substitution 897
38.3
0
dx
(4+x2)3
39. √x2 −4
40. √x
x2 −1 dx
41.2
2
√
3
√
x2 −1
42. 2x−3
√
9−4x2 dx
43. x
4−x2 dx
44 Find √
k2− x2dx for any constant k.
45 Show that the area enclosed by the ellipse x a22 +y2
b2 = 1, where a and b are positive constants, is given by π ab.
y b
–b
x2
a2
y2
b2 + = 1
Hint: Find the area in the first quadrant and multiply by 4 Express y as a function of x Notice the similarity between the formula for the area inside a circle and the formula for the area inside an ellipse.
46 Find
sec3x dx as follows: Use integration by parts with u = sec x The resulting
new integral will contain tan2x Replace tan2xby sec2x− 1 and split the integral into the difference of two integrals,
sec3x dx− sec x dx Integrate the latter and solve
algebraically for the former
47 Derive the reduction formula
tann x dx=tann−1x
n− 1 − tann−2x dx,
where n is an integer greater than or equal to 2 (To do this, rewrite the integrand as
tann−2x· tan2x and use a Pythagorean identity to convert tan2x into an expression involving sec x.)
48 Derive the formula sin Ax sin Bx=1
2[cos(A − B)x − cos(A + B)x] given in this section (Begin with the addition formula for cosine.)
Trang 829.3 INTEGRATION USING PARTIAL FRACTIONS
Practical motivation: Models commonly used in epidemiology and population biology are
often based on what is known as a logistic growth model For instance, while population growth may initially look exponential, due to limited resources growth generally levels off
at what is known as the carrying capacity for the population In order to derive an expression
for P (t), the population at time t, one computes 1
kP (P −L) dP , where L is the carrying capacity and k is a constant determined by the population The methods of this section
enable us to calculate this integral
Suppose an integral is of the form P (x)
Q(x) dx where P (x) and Q(x) are polynomials and Q(x) factors The method of partial fractions is an algebraic technique enabling us to
split up the integrand into the sum of simpler rational functions For example, we know that
1
x+ 2 −
1
x+ 3 =
x + 3 − (x + 2) (x + 2)(x + 3) =
1
x2+ 5x + 6.
Suppose we are struggling to find 1
x2+5x+6 dx If we replace the integrand by the
equiv-alent expression x+21 − 1
x+3, our struggles are over.
1
x2+ 5x + 6 dx=
1
x+ 2 −
1
x+ 3
dx
=
1
x+ 2 dx−
1
x+ 3 dx
= ln |x + 2| − ln |x + 3| + C
= ln
x x+ 2+ 3 + C
The method of decomposition into partial fractions is a method of integrating rational
functions, functions of the form P (x) Q(x) , where P (x) and Q(x) are polynomials by
decom-posingP (x) Q(x) into the sum of simpler rational functions
The method of partial fraction decomposition works only if P (x) Q(x) is a proper fraction,
that is, only if the degree of P is strictly less than the degree of Q(x): deg(P ) < deg(Q) Suppose deg(P ) ≥ deg(Q); the fraction is improper Then we can do long division to split
the fraction into a sum of a polynomial and a proper rational function We can use the method
of partial fractions on the proper part We illustrate this below
◆ EXAMPLE 29.17 Integrate x2
x2 −1 dx.
x2
x2− 1 = 1 +
1
x2− 1 or 1+
1
(x − 1)(x + 1).
Trang 929.3 Integration Using Partial Fractions 899 1
(x −1)(x+1) can be decomposed into a sum of the form x A−1 + B
x+1 where A and B are constants
1
(x − 1)(x + 1)=
A
x− 1+
B
x+ 1 Clear denominators.
1= A(x + 1) + B(x − 1) Gather x terms; gather constants 0x + 1 = (A + B)x + (A − B)
This can be true only if A + B = 0 and A − B = 1 Solve these simultaneous equations to obtain A=1
2and B= −1
2 Then
x2
x2− 1= 1 +
1 2
1
x− 1−
1 2
1
x+ 1.
x2
x2− 1dx=
1+1 2
1
x− 1 −
1 2
1
x+ 1
dx
= x + 1
2 ln|x − 1| −1
2ln|x + 1| + C ◆ The first step in partial fraction decomposition of a proper rational function is factoring the denominator It can be proven that any polynomial can be factored into the product of
linear and irreducible quadratic factors A quadratic ax2+ bx + c is irreducible if it cannot
be factored into the product of two linear factors, that is, if the discriminant, b2− 4ac, is
negative We’ll look at polynomials that are not hard to factor Computer algebra systems can take care of more difficult factoring jobs
A proper rational function can be decomposed into the sum of proper rational functions We’ll look at some examples before giving the general procedure Suppose the denominator factors into the product of distinct linear factors For example, consider −3
(x +1)(x−2) We can
express −3
(x +1)(x−2) as a sum of the form x A+1 + B
x−2 The task of determining A and B reduces to simultaneous equations
−3
(x + 1)(x − 2)=
A
x+ 1+
B (x − 2) Get a common denominator.
−3
(x + 1)(x − 2)=
Ax − 2A + Bx + B (x + 1)(x − 2) Equate the numerators.(Equivalently, clear denominators.)
−3 = Ax − 2A + Bx + B Collect like powers of x.
−3 = (A + B)x + (B − 2A) 0x − 3 = (A + B)x + (B − 2A) Equate the coefficients of x;
equate the constant terms
Then
A + B = 0
−3 = B − 2A. simultaneous linear equations
A = −B so − 3 = B − 2(−B)
− 3 = 3B
− 1 = B and A = 1,
(x + 1)(x − 2)=
1
x+ 1−
1
x− 2. Check by adding.
Trang 10Knowing the partial fraction decomposition of −3
x2−x−2 makes finding its antiderivative a
snap
x2− x − 2 dx=
1
x+ 1−
1
x− 2dx
= ln |x + 1| − ln |x − 2| + C
= ln
x+ 1
x− 2
+ C
There’s a shortcut to finding A and B Given −3
(x +1)(x−2)= A
x−2, clear denominators
and then evaluate both sides at values of x that would have made the denominator zero It
can be shown that this is legal; certainly it is quick
− 3 = A(x − 2) + B(x + 1) Evaluate at x = −1: − 3 = A(−1 − 2) ⇒ A = 1 Evaluate at x = 2: − 3 = B(2 + 1) ⇒ B = −1
We’ll apply this shortcut in the next example
x2−3x2+2x dx.
SOLUTION The integrand is a proper fraction, so we factor the denominator
x2+ 1
x3− 3x2+ 2x =
x2+ 1
x(x2− 3x + 2)=
x2+ 1
x(x − 2)(x − 1)
The denominator has distinct linear factors, so we set up partial fractions as follows
x2+ 1
x(x − 2)(x − 1)=
A
x− 2 +
C
x− 1
x2+ 1 = A(x − 2)(x − 1) + Bx(x − 1) + Cx(x − 2) Evaluate at x = 0: 1 = A(−2)(−1) ⇒ A =1
2
Evaluate at x = 2: 5 = B(2)(1) ⇒ B =5
2
Evaluate at x = 1: 2 = C(1)(−1) ⇒ C = −2
x2+ 1
x3− 3x2+ 2x = −
1 2
x +
5 2
x− 2 −
2
x− 1 Therefore
x2+ 1
x3− 3x2+ 2x dx=
1 2
1
x dx+5
2
1
x− 2 dx− 2
1
x− 1 dx
=1
2 ln|x| +5
2ln|x − 2| − 2 ln |x − 1| + C ◆