Download Class Exercise Introduction

BGU Physics Dept. Introduction to Mathematical Methods in Physics
Class Exercise
Integrals
Introduction
Integration by parts:
For integral of the sort
Z b
f (x)g(x)dx ,
a
introduce two intermediary functions U(x) and V(x)
U = f (x)
,
dV = g(x)dx
and find
dU =Rf 0 (x)dx
,
V = g(x)dx
to use the formula
Z
b
U dV = (U V
a
)|ba
Z
−
b
V dU
a
Polynomial over Polynomial
For integral of the sort
Z
P (x)
dx ,
Q(x)
where P(x) and Q(x) are polynomials, we solve in few steps:
I If deg[F ] > deg[G], then it is necessary to perform the division
P (x)
R(x)
= D(x) +
,
Q(x)
Q(x)
via polynomial long division or otherwise. D(x) is now a simple polynomial which we can easily
integrate and we left with
Z
R(x)
dx ,
Q(x)
which by definition has deg[R] < deg[Q].
II Write Q(x) as multiplication of irreducible factors. (e.g. x2 − 4 = (x − 4)(x + 4))
III Use partial fractions method to simplify the integral
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• The thumb rule is that each nominator N (x) should be a polynomial with one degree
below the order of the denominator, e.g.:
1
a
bx + c
d
= + 2
+
x(3x2 + 5)(2x + 5)
x 3x + 5 2x + 5
(1)
• If one of the factors is of the sort (x − α)−r than we write has r partial fractions with
constant nominator and a decreasing power dwnominator e.g.:
3x + 5
a
b
=
+
(2x − 1)2
(2x − 1)2 (2x − 1)
IV Solve the integral. Some useful integrals:
Z
1
= ln |x − a| + C
x−a
Z
1
1
= arctan xa + C
x2 + a2
a
Z
x
1
= ln(x2 + a2 )
x2 + a2
2
Trigonometric integrals
For integrals composed from trigonometric functions R(sin x, cos x)
• If R(− sin x, cos x) = −R(sin x, cos x) use t = cos x.
• If R(sin x, − cos x) = −R(sin x, cos x) use t = sin x.
• If R(− sin x, − cos x) = R(sin x, cos x) use t = tan x or t = cot x.
• If nothing else works we can use Universal Trigonometric Substitution (UTS) tan
that
1 − t2
1 + t2
2t
sin(x) =
1 + t2
2
dx =
dt
1 + t2
cos(x) =
Exrecises
1. Solve
R
(a) tan xdx
R
3 +1
(b) m4m+4m+1
dm
R dx
(c) x lnn (x) dx
Solution
2
x
2
= t so
(a)
Z
Z
sin x
dx =
cos x
tan xdx =
Z
dt
cos x = t
=−
− sin xdx = dt
t
= − ln t + C = − ln(cos x) + C
(b)
Z
m3 + 1
dm =
m4 + 4m + 1
=
m4 + 4m + 1 = t
(m3 + 1)dm = dt
4
Z
=
dt
1
= ln t + C
4t
4
1
ln(m4 + 4m + 1) + C
4
(c)
Z
dx
dx =
x lnn (x)
=
ln(x) = t
dx
x = dt
Z
=
dt
t1−n
=
=
tn
1−n
ln1−n (x)
1−n
2. Solve
R
dθ
(a) 1+sin(θ)
R
(b) sin5 (u) cos2 (u)du
Solution
(a) First Way:
Z
Z
2dt
θ
1
UTS: tan
=t =
2t 1 + t2
2
1 + 1+t2
Z
Z
2dt
2dt
2
2
=
=
=−
+C = −
2
2
1 + 2t + t
(1 + t)
1+t
1 + tan
dθ
=
1 + sin(θ)
θ
2
+C
Second Way:
Z
Z
Z
Z
1
dθ
1 − sin θ
sin θ
dθ =
=
dθ −
dθ
2
2
1 + sin(θ)
cos θ
cos2 θ
1 − sin θ
Z
sin θ
cos θ = t
dθ =
= tan θ −
− sin θdθ = dt
cos2 θ
Z
dt
1
1
= tan θ +
= tan θ − + C = tan θ −
+C
t2
t
cos θ
which is same as the first up to a constant
(b)
Z
5
Z
2
2
2
sin(u)(1 − cos (u)) cos (u)du =
sin (u) cos (u)du =
Z
=
2
2 2 2
(1 − z ) z dz =
3
= cos (u)
Z
cos(u) = z
− sin(u)du = dz
z 2 − 2z 4 + z 6 dz =
1 2
1
− cos2 (u) + cos4 (u) + C
3 5
7
3
z3
z5 z7
−2 +
+C
3
5
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3. Solve
R
(a) ln(x)dx
R
(b) x2 sin(x)dx
Solution
(a)
Z
U = ln(x) dU = dx
x
ln(x)dx =
dV = dx
V =x
Z
= x ln(x) − dx = x ln(x) − x
Z
x2 sin(x)dx =
(b)
Z
U = x2
dU = 2xdx
= −x2 cos(x) + 2 x cos(x)dx
dV = sin(x)dx V = − cos(x)
Z
U =x
dU = dx
2
=
= −x cos(x) + 2x sin(x) − 2 sin(x)dx
dV = cos(x)dx V = sin(x)
= (2 − x2 ) cos(x) + 2x sin(x)
4. Solve
ex cos x
R
Solution
Z
x
x
Z
e cos x = e cos x +
Z
5. Solve
R
(a)
R
(b)
ex cos x =
x
x
x
e sin x = e cos x + e sin x −
Z
ex cos x
ex (cos x + sin x)
2
4x2 −13x+13
dx
(x−1)(x−2)2
x
dx
x2 −3x+2
Solution
(a) First we decompose to partial fractions:
4x2 −13x+13
(x−1)(x−2)2
A
B
C
= (x−1)
+ (x−2)
+ (x−2)
2
4x2 − 13x + 13 = A(x − 2)2 + B(x − 1)(x − 2) + C(x − 1)
4x2 − 13x + 13 = (A +
B)x2 + (−4A − 3B + C)x + (4A + 2B − C)
4=A+B

⇒ −4A − 3B + C = −13

4A + 2B − C = 13
A = 4; B = 0; C = 3;
Now we can solve:
Z
Z
Z
4x2 − 13x + 13
4
3
dx
=
dx
+
dx
2
(x − 1)(x − 2)
x−1
(x − 2)2
= 4 ln |x − 1| −
4
3
x−2
(b)
Z
x
dx =
2
x − 3x + 2
Z
x
dx
(x − 1)(x − 2)
x
(x−1)(x−2)

B
+ (x−2)

=
1=A+B
x = A(x − 2) + B(x − 1) →
⇒ A = −1; B = 2
0 = −2A − B
Z 2
1
=
dx = 2 ln |x − 2| − ln |x − 1| + C
−
(x − 2) (x − 1)


=
A
(x−1)
if any typos or corrections found please let me know: Ben Yellin [email protected]
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