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Math 142, Exam 1 Information.
9/14/10, LC 412, 9:30 - 10:45.
Exam 1 will be based on:
• Sections 7.1 - 7.5.
• The corresponding assigned homework problems
(see http://www.math.sc.edu/∼boylan/SCCourses/142Fa10/142.html)
At minimum, you need to understand how to do the homework problems.
• Lecture notes: 8/19 - 9/8.
Topic List (not necessarily comprehensive):
You will need to know: theorems, results, and definitions from class.
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§7.1: Integration
by parts.
To integrate f (x) dx using the partsR method, identify u and
R
R
dv so that f (x) dx = u dv. To obtain v from dv, integrate: v = dv. The parts formula
is then given by
Z
Z
u dv = uv −
v du.
• You may have to apply parts two or more times.
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Example: x2 cos x dx.
• Repeated iteration of parts may produce the original integral, I. Treating I as a variable,
this gives a linear equation in the variable I. To evaluate the original integral solve the
linear equation for I.
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Example: ex cos x dx.
• Further examples: Suppose that f (x) =
– xj (trig. function) (ex. f (x) = x sin 4x),
– xj (exponential function) (ex. f (x) = x3 e2x ), or
– xj (logarithm function) (ex. f (x) = x ln x).
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Then it is reasonable to try to use parts to integrate f (x) dx.
§7.2: Trigonometric integrals.
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• I = sinm x cosn x dx. It is helpful to recall that
sin2 x + cos2 x = 1,
u = sin x =⇒ du = cos x dx,
u = cos x =⇒ du = − sin x dx.
(i) Suppose that m = 2j + 1 is odd. Let u = cos x; then du = − sin x dx.
We have
Z
Z
Z
2
j
n
2
j
n
I = (sin x) cos x(sin x dx) = (1−cos x) cos x(sin x dx) = − (1−u2 )j un du.
Point:
Since we have an odd power of sin, we can rewrite the integral as
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(function of cos x)(sin x dx).
(ii) Suppose that n = 2k + 1 is odd. Let u = sin x; then du = cos x dx.
We have
Z
Z
Z
m
2
k
m
2
k
I = sin x(cos x) (cos x dx) = sin x(1−sin x) (cos x dx) = um (1−u2 )k du.
Point:
Since we have an odd power of cos, we can rewrite the integral as
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(function of sin x)(cos x dx).
(iii) Suppose that neither m nor n are odd (both are even). In this case apply a suitable
trigonometric identity such as one of the following:
cos 2x = 1 − sin2 x = 2 cos2 x − 1 =⇒ sin2 x =
sin 2x = 2 sin x cos x =⇒
1 − cos 2x
1 + cos 2x
, cos2 x =
;
2
2
1 2
sin 2x = sin2 x cos2 x.
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Example:
Z
Z
Z 1
1 + cos 2t
2
4
2
2
2
2
sin t cos t dt = sin t cos t cos t dt =
· sin 2t ·
dt
4
2
Z sin2 2t cos 2t
1
2
sin 2t +
=
dt
4
2
Z Z
1
1 − cos 4t
1
=
dt +
sin2 2t cos 2t dt.
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2
8
2
• I=
R
tanm x secn x dx. It is helpful to recall that
tan2 x + 1 = sec2 x,
u = tan x =⇒ du = sec2 x dx,
u = sec x =⇒ du = sec x tan x dx.
(i) Suppose that n = 2k > 0 is even. Let u = tan x; then du = sec2 x dx. We have
Z
Z
m
2
k−1
2
I = tan x(sec x) (sec x dx) = tanm (tan2 x + 1)k−1 (sec2 x dx)
Z
= um (u2 + 1)k−1 du.
Point:
Since we have an even power of sec, we can rewrite the integral as
R
(function of tan x)(sec2 x dx).
(ii) Suppose that n ≥ 1 and m = 2j + 1 ≥ 1 is odd. Let u = sec x;
then du = sec x tan x dx. We have
Z
Z
2
j
n−1
I = (tan x) sec
x(sec x tan x dx) = (sec2 x − 1)j secn−1 x(sec x tan x dx)
Z
= (u2 − 1)j un−1 du.
Point: Since we have Ran odd power of tan and at least one power of sec, we can
rewrite the integral as (function of sec x)(sec x tan x dx).
(iii) If you cannot rewrite the integral as either
Z
Z
2
(function of tan x)(sec x dx) or
(function of sec x)(sec x tan x dx),
then you must use other methods and facts. For example, the formulas
Z
Z
tan x dx = ln | sec x| + C,
sec x dx = ln | sec x + tan x| + C
are useful. For a second example, one uses parts to compute
R
sec3 x dx.
• If the integrand does not contain the natural pairings (sin x, cos x), (tan x, sec x), or
(cot x, csc x), it may be possible to use an identity to rewrite the integrand in terms of
these pairings.
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R
Example: tan x cos2 x dx = sin x cos x dx.
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R
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• I = sin ax cos bx dx, sin ax sin bx dx, cos ax cos bx dx. One can add and subtract
sin(x ± y) = sin x cos y ± sin y cos x, cos(x ± y) = cos x cos y ∓ sin x sin y
to obtain
1
1
(sin(A − B) + sin(A + B)) , sin A sin B = (cos(A − B) − cos(A + B)) ,
2
2
1
cos A cos B = (cos(A − B) + cos(A + B)) .
2
3
sin A cos B =
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§7.3: Trigonometric substitution. Compute I = f (x) dx. In each of the following cases
(with a > 0), we make a substitution involving the variable θ. For this purpose, we draw a
right triangle with base angle theta.
√
• Suppose that f (x) contains a2 − x2 .
(i) The right triangle has
∗ hypotenuse a;
∗ side opposite θ equal to x;
√
∗ side adjacent to θ equal to a2 − x2 .
(ii) The substitution is sin θ = x/a, with
−π/2 ≤ θ ≤ π/2. Then a sin θ = x, so
√
√
a2 −x2
dx = a cos θ dθ. Moreover, cos θ = a , so a cos θ = a2 − x2 .
• Suppose that f (x) contains
√
a2 + x 2 .
(i) The right triangle has
√
∗ hypotenuse a2 + x2 ;
∗ side opposite θ equal to x;
∗ side adjacent to θ equal to a.
(ii) The substitution is tan θ = x/a, with
−π/2 < θ < π/2. Then a tan θ = x, so
√
√
a2 +x2
2
dx = a sec θ dθ. Moreover, sec θ = a , so a sec θ = a2 + x2 .
• Suppose that f (x) contains
√
x2 − a2 .
(i) The right triangle has
∗ hypotenuse x;
√
∗ side opposite θ equal to x2 − a2 ;
∗ side adjacent to θ equal to a.
(ii) The substitution is sec θ = x/a with 0 ≤ θ < π/2 or √π ≤ θ < 3π/2. Then
2
2
a sec θ = x, so dx = a sec θ tan θ dθ. Moreover, tan θ = x a−a ,
√
so a tan θ = x2 − a2 .
√
More generally, suppose that f (x) contains rx2 + sx + t. To use the above substitutions, complete the square to convert rx2 + sx + t to the form x2 ± a or a2 − x2 .
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§7.4: Partial fractions.
• Q(x) factors as a product of distinct, linear factors: Q(x) = (a1 x + b1 ) · · · (aj x + bj ).
A1
Aj
P (x)
=
+ ··· +
.
Q(x)
a1 x + b + 1
aj x + b j
(ii) Clear denominators, and solve for A1 , . . . Aj . This can be done several ways:
∗ Compare coefficients of powers of x, giving a linear system of equations.
∗ Substitute specific values for x; specifically, choose values of x which have
Q(x) = 0.
(i) Write
• Q(x) factors as a product of linear factors, some of which are repeated.
(i) Corresponding to a repeated factor (ax + b)r of Q(x), we have the terms
A1
A2
Ar
+
+ ···
2
ax + b (ax + b)
(ax + b)r
in the partial fraction decomposition of P (x)/Q(x).
(ii) Clear denominators and solve for the constants in the decomposition as before.
• Q(x) contains irreducible quadratic factors, none of which is repeated.
(i) Suppose that Q(x) has a factor ax2 + bx + c with b2 − 4ac < 0 (so that it is
irreducible). Corresponding to this factor, we have the term
Ax + B
ax2 + bx + c
in the partial decomposition of P (x)/Q(x).
(ii) Clear denominators and solve for constants as before.
(iii) The integral
Z
dx
1
−1 x
=
tan
+C
x 2 + a2
a
a
may be useful.
• Q(x) contains a repeated irreducible quadratic factor.
(i) Suppose that Q(x) has the factor (ax2 + bx + c)r with b2 − 4ac < 0 (so that
ax2 + bx + c is irreducible). Corresponding to this factor, we have the term
A1 x + B1
A2 x + B2
Ar x + Br
+·+
+
2
2
2
ax + bx + c (ax + bx + c)
(ax2 + bx + c)r
in the partial fraction decomposition of P (x)/Q(x).
(ii) We clear denominators and solve for the constants as before.
p
n
g(x), it may be reasonable to try
Rationalizing substitutions.
If
an
integrand
contains
p
n
the substitution u = g(x) (if all else fails).
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§7.5: Integration strategy.
Other useful integrals (in addition to those above):
Z
ax
+ C,
ax dx =
ln a
Z
x
dx
√
= sin−1
+ C.
a
a2 − x 2
Do not forget the following:
• Simplify the integrand if possible.
• Look for obvious u-substitutions.
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