1 Integration methods

Students should have no difficulties in solving the problems in this document after taking math 222, a prerequisite course for physics 202.
Physics 201 and 202 both contain line integrals, and simple double or even triple integrals (but this latter case is rare).
This document illustrates several important points; it details what students should know coming into 201/202, or what they will be expected to learn in 201/202. Students must prepare themselves to gain some level of mastery of these concepts in 201/202, either through prior preparation in math courses or through hard work (homework and problem solving) and participation in class.

1.1 Integration by parts

This technique is used to rewrite integrals of products of two functions. Consider functions u(x) and v(x) and integrals of the form

∫ ∫ uf u(xf ) du(x)--- u v du = u(x ) v(x) dx = C + D 0 0 dx

which is numerically equal to the area of the regions C and D in the ﬁgure.

Notice that the shaded region has area

uf vf - u0v0 = A + B + C + D

Using this our original integral can be rewritten as

∫ u ∫ v f v du = u(x) v(x) ∣f - fu dv = (A + B + C + D) - (A + B) u0 0 v0

which is the entire integration by parts scheme.

Example

Consider integration by parts for

∫ π 2 I = 0 x sin x dx

We let v = x² and u = - cos(x) to get after one application

2 π ∫ π I = - x cos x ∣0 + 2x cos x dx 0

and we apply again to the last integral with v = 2x and u = sinx to get

2 π π ∫ π I = - x cos x ∣0 + 2x sin x ∣0 - 2 cos x dx 0

The last integral is elementary;

2 π π π I = - x cos x ∣0 + 2x sin x ∣0 - 2 sin x ∣0

A common situation is to have integration by parts apparently take you around in circles;

Example

apply partial integration to

∫ I = ∞ e- ax sin x dx 0

with v = e^-ax and u = - cos x;

∫ - ax ∞ ∞ - ax I = - e cos x ∣0 - a 0 e cos x dx

and now integrate the last integral by parts again with v = e^-ax and u = sin x;

∫ ∞ I = 1 - ae - ax sin x∣∞ - a2 e- ax sin x dx = 1 - a2I 0 0

and so

∫ ∞ - ax 1 I = e sin x dx = --------- 0 1 + a2

This is a useful integral, it is the Laplace transform of a sine function, and so will appear in classical mechanics and electronics in the Newtonian equations of motion for a particle with a sinusoidal (in time) force applied to it, or for the current in a circuit with a sinusoidal applied voltage.

1.2 Variable Changes

Some of the simplest and most useful are;
1.

∫ -x-dx---- 2 x + 2

Let y = x² + 2, then dy = 2xdx so that xdx = dy
2

and our integral can be written as

∫ x dx 1 ∫ dy 1 1 --------- = -- ---- = -- ln ∣y∣ + C = --ln ∣x2 + 2 ∣ + C x2 + 2 2 y 2 2

∫ dx √----------- 1 - x2

Let x = sin θ, then dx = cos θ dθ and √ -------
1 - x2

= cos θ and our integral becomes

∫ ∫ √--dx------- - 1 2 = dθ = θ + C = sin x + C 1 - x

which gives us a rather nice deﬁnition of the arcsine or inverse-sine function.

∫ dx --------- x2 + 1

Let x = tan θ, then dx = codθs2-θ

(prove this, it is a nice exercise), and 1 + x² = 1 + 2
scinos2θθ-

and our integral becomes

∫ dx ∫ ---------= d θ = θ + C = tan - 1 x + C x2 + 1

which, again, provides us with a nifty representation of an inverse trig function.

Example

Determine a power series expansion for tan ^-1x valid near x = 0.
We can use the previous problem

∫ x --dy----- - 1 0 2 = tan x y + 1

and if we recognize that in this expression 0 ≤ y ≤ x and x is very small, then y is very small and we can replace 1+1y2

with its power series for y ≈ 0 in the integrand;

1 ---------= 1 - y2 + y4 - y6 + ⋅ ⋅ ⋅ 1 + y2

which you should prove. We now integrate term by term;

∫ 3 5 7 x 2 4 6 x--- x--- x--- 0 (1 - y + y - y + ⋅ ⋅ ⋅) dy = x - 3 + 5 - 7 + ⋅ ⋅ ⋅

and so very near x = 0 we ﬁnd that

3 5 7 - 1 x x x tan x = x - ----+ ----- ----+ ⋅ ⋅ ⋅ 3 5 7

∫ √---dx------ 2 1 + x

There is a standard substitution to use whenever your integrand has a radical containing 1 ± x²; try x = cot θ. Then dx = - sdinθ2θ

and 1 + x² = si1n2θ

and the integral becomes

∫ ∫ ∫ √---dx------= - -d-θ-- = - ------d-θ------- 2 sin θ θ- θ- 1 + x 2 sin 2 cos 2

At this stage we notice (after we have become sufficiently experienced) that

-d-- θ- 1----1------1---- -------1-------- ln tan = 2 θ- θ-= θ- θ- d θ 2 2 cos 2 tan 2 2 sin 2 cos 2

and so the integrand is the derivative of something;

∫ ---dx------- ∫ -d-θ-- θ- √ --------- = - = - ln tan + C 1 + x2 sin θ 2

We now need to recover the answer in terms of x, but

1 2 tan θ- x = cot θ, -- = tan θ = ----------2--- x 1 - tan2 θ- 2

which we solve for

2 θ- θ- tan + 2x tan - 1 = 0 2 2

θ √ --------- tan --= - x ± x2 + 1 2

and so

∫ √ --------- √ --------- √---dx------= - ln ∣ - x ± x2 + 1∣ = ln ∣x ± x2 + 1 ∣ 2 1 + x

since

√ --------- √ --------- 1 x ± x2 + 1 x ± x2 + 1 √ --------- ---------√---------- = --------√--------------------√------------ = --------------------- = x ± x2 + 1 - x ± x2 + 1 (x ± x2 + 1)( - x ± x2 + 1) - x2 + (x2 + 1)

This was a rather complicated example, however once an integral has been done, you may archive it in your own personal integral table and re-use the result without having to go through all of the trouble of rederiving it.

∫ dx ∘------------- (x2 + 1)3

is an integral that will come up once or twice in Physics 202. We use the x = tan θ variable change on it, to get

∫ ∫ ∘----dx------- 2 3 = cos θ d θ = sin θ + C (x + 1)

x = √-----------+ C x2 + 1

using the triangle ﬁgure to resolve the trig functions.

∫ √ --------- 1 - x2 dx

This integral comes up if we want to compute the area of a unit circle, or at least the part of the circle that lies in the ﬁrst quadrant. We use x = sin θ to get

∫ √ --------- ∫ 1 - x2 dx = cos2 θ d θ

which we simplify by applying a trigonometric identity

∫ ∫ 2 1- 1- 1- cos θ dθ = (1 + cos 2θ) d θ = (θ + sin 2 θ) 2 2 2

and in terms of x, cos θ = √ -------
1 - x2

, so

∫ √ --------- √ --------- 2 1- - 1 1- 2 1 - x dx = (sin x + 2x 1 - x ) + C 2 2

1.3 Partial Fractions

This method is used when one must evaluate integrals of the form

∫ ---------------dx----------------- (x - a)(x - b)(x - c) ⋅ ⋅ ⋅

It is very easy to show that if we suppose that a function of this form can be expanded as a sum such as

1 a a a f (x) = ------------------------------= -----1-- + ----2--- + ----3--- (x - a)(x - b)(x - c) x - a x - b x - c

then we can ﬁnd the coefficients a_i by a limit taking process

x-----a- x-----a- a1 = lxi→ma(x - a)f (x) = lxi→ma (a1 + a2 + a3 ) x - b x - c

as long as b

a and c

a. Then the last two terms will be zero in the limit. Such a decomposition is very powerful and is called a partial fraction or pole decomposition and the limit is called the residue

1 Res a Res b Res c f (x) = ------------------------------= -------- + -------- + -------- (x - a)(x - b)(x - c) x - a x - b x - c

--------1---------- Res a = lxi→ma(x - a)f (x) = (a - b)(a - c)

and so forth, we ﬁnd that

-------------1-------------- b-----c- a-----c- a-----b- f (x) = [ - + ] (a - b)(b - c)(a - c) x - a x - b x - c

∫ dx --------- x2 - 1

We perform the pole expansion

1 1 1 1 ---------= --(--------- -------) x2 - 1 2 x - 1 x + 1

and integrate directly

∫ dx 1 ∫ dx dx 1 1 x - 1 ---------= -- (--------- -------) = --(ln ∣x - 1∣- ln ∣x+1 ∣) = --ln ∣--------∣ x2 - 1 2 x - 1 x + 1 2 2 x + 1

∫ dx --------- x2 + x

This can be expanded as

∫ dx ∫ 1 1 x --------- = dx( -- - -------) = ln ∣x ∣- ln ∣x + 1∣+ C = ln ∣-------∣+ C x2 + x x x + 1 x + 1

∫ ----dx------- 2 x(x + 1)

The residue method requires a modiﬁcation in order to be able to handle multiple roots, in this case the required decomposition is

1 1 x + 2 1 x + 1 1 ------------- = -- - ----------- = -- - ----------- - ----------- x(x + 1)2 x (x + 1)2 x (x + 1)2 (x + 1)2

and so

∫ -----dx------ ---1---- 2 = ln ∣x ∣ - ln ∣x + 1∣ + + C x(x + 1) x + 1

If we have an expression in the integrand such as

1 f (x) = -------------------n- (x - a)(x - b)

we propose an expansion

----------1---------- A1(a)--- A1(b)--- --A2(b)---- --An(b)---- n = + + 2 + ⋅ ⋅ ⋅ + n (x - a)(x - b) x - a x - b (x - b) (x - b)

Notice that

lim (x - a)f (x) = A1(a) x→a

and

lim (x - b)nf (x) = An(b) x→a

but the other coefficients are harder to ﬁnd. To get all of the others, compute

n n (x-----b)--A1(a)--- ∑n n- m (x - b) f(x) = + Am(b)(x - b) x - a m=1

Notice that

d n lim ----[(x - b) f (x)] = An - 1(b) x→b dx

d2 lim ----[(x - b)nf (x)] = 2An - 2(b) x →b dx2

3 -d--- n lxim→b dx3 [(x - b) f (x)] = 3 ⋅ 2An - 3(b)

which you really should verify by direct calculation. This allows us to compute all of the other residues A_m(b) and ﬁnish the expansion. If you go on in math and take a course in complex variables, you will learn that the residue technique together with Cauchy’s Theorem allow you to compute virtually any deﬁnite integral in a straightforward and systematic way. In fact if you limit your attention to real variables only, there are no universal procedures for evaluating integrals, only a bag of tricks. However in complex variables there are genuine procedures that allow for relatively painless evaluation of even the most ferocious deﬁnite integrals.

10.

∫ ---------dx----------- 2 2 (x - a) (x - b)

∫ 1 1 1 1 - 2 1 - 2 1 = dx ( ----------------------+ ----------------------+ ----------- ----------+ ---------------------) (a - b)2 (x - a)2 (b - a)2 (x - b)2 (a - b)3 (x - a) (b - a)3 (x - b)

and all of these integrals are elementary.

All of the techniques presented so far have been elementary, though some of the results of their application have not. You will ﬁnd that the residue method has very many applications, not only to the evaluation of limits and integrals, but also the the construction of very powerful representations of ill-behaved functions. A beautiful example is the cot x function. We have already seen that near x = 0, it behaves badly,

1- x- cot x = - + ⋅ ⋅ ⋅ x 3

We may suspect that since it has a sin x function in the denominator, it probably behaves just as badly near all of the other zeroes of the sine function, namely x = kπ, k = 0,±1,±2, ⋅⋅⋅

. We might then propose a rather odd series

a0- --a1---- -a---1-- ---a2----- --a---2--- cot x = + + + + + ⋅ ⋅ ⋅ x x - π x + π x - 2 π x + 2π

We can compute all of the coefficients by residue or limit methods, since

lim x cot x = 1 = a0 x →0

lim (x - π) cot x = 1 = a x → π 1

xl→imn π(x - n π) cot x = 1 = an

the evaluation of these limits would be a good exercise for the reader. We obtain the remarkable expansion

∞∑ 1 1 ∞∑ 2x cot x = ---------- = -- + -------------- k= - ∞ x - kπ x k=1 x2 - k2 π2

which you will probably never use, but can appreciate how easy it was to obtain using only the most common tools of calculus (limits) and how beautiful it is!

11. A new category of transcendental functions have been invented to allow one to evalute integrals of types

∫ ∫ ∫ √---dx------ --dx----- √---dx------ 2, 2, 2 1 + x 1 - x x - 1

called hyperbolic trig functions sinh x, tanh x and cosh x. Consider the function f(x) that satisﬁes the equation

2 d--f-(x)- ′ 2 = f (x), f (0) = 1, f (0) = 0 dx

We can construct a series for it valid near x = 0 since we know all of its derivatives at x = 0;

′′ ′′′ ′ f (0) = f (0) = 1, f (0) = f (0) = 0, ⋅ ⋅ ⋅

and so

n ∞∑ -1-d--f-(0)- n 1- 2 -1-- 4 f (x) = n (x - 0) = 1 + x + x + ⋅ ⋅ ⋅ n=0 n! dx 2 24

which is handy. Now multiply the original differential equation by f′(x)

′′ ′ ′ f (x) f (x) = f (x) f (x)

but now both sides are the derivatives of things

d 1 ′ ′ d 1 2 ----(--f (x) f (x)) = -----f (x) dx 2 dx 2

and can therefore be integrated

∫ ∫ x -d-- 1- ′ ′ 1- ′ ′ ′ ′ x -d--1- 2 1- 2 2 0 ( f (x) f (x)) dx = (f (x)f (x) - f (0)f (0)) = 0 f (x)dx = (f (x) - f (0)) dx 2 2 dx 2 2

∘ ------------- ′ ′ 2 df-(x)-- 2 f (x) f (x) = f (x) - 1, = f (x) - 1 dx

resulting in

∫ df ∫ √----------- = dx = x f 2 - 1

and so the inverse function of f(x), namely f^-1(x), is the solution to our third new integral above. This is called the cosh(x) or hyperbolic cosine function. Notice that since

2 4 6 x--- x--- x---- cosh(x) = 1 + + + + ⋅ ⋅ ⋅ 2 24 720

and

2 3 4 5 x x--- x--- x--- x---- e = 1 + x + + + + + ⋅ ⋅ ⋅ 2 6 24 120

and

2 3 4 5 - x x--- x--- x--- -x--- e = 1 - x + - + - + ⋅ ⋅ ⋅ 2 6 24 120

and that the new function can be written in terms of our exponentials

x - x e---+--e---- cosh(x) = 2

There is a solution to

2 d--g(x)-- ′ 2 = g(x), g(0) = 0, g (0) = 1 dx

whose series is

x3 x5 g(x) = x + ----+ ----- + ⋅ ⋅ ⋅ 6 120

that has the following interesting property; differentiate this to get

3 d--g(x)-- dg(x)--- 3 = dx dx

and deﬁne

dg(x)--- h(x) = dx

but then note that h(0) = g′(0) = 1, and h′(0) = g′′(0) = g(0) = 0, and so h(x) = f(x) = cosh(x)! This function is called the hyperbolic sine

g(x) = sinh(x)

and since we know then that f′(x) = h′(x) = g′′(x) = g(x) and

f ′(x) f ′(x) = f 2(x) - 1,

we obtain the fundamental identity for hyperbolic functions

cosh2(x) - sinh2(x) = 1

analogous to cos ²(x) + sin ²(x) = 1 for the trig (circular) functions. Now we can perform

∫ dx ∫ dx ---------, √----------- 1 + x2 x2 - 1

For the last one, let x = cosh y, then dx = sinh y dy and

∫ dx ∫ sinh y dy √-----------= ------------ = y + C 2 sinh y x - 1

and so

∫ dx √-----------= cosh - 1 x + C x2 - 1

similarly

∫ dx √-----------= sinh - 1 x + C 1 + x2

sub

2 Problems

1. Use the method illustrated in this section to prove that

k 1- ∞∑ --(---1)------ csc x = + 2x 2 2 2 x k=1 x - k π

2. Show that

x3 3x5 15x7 sin - 1 x = x + ----+ -----+ ------- + ⋅ ⋅ ⋅ 6 40 336

by integrating term by term an integral expression for arcsine of x.

3. Prove the following integrations

∫ dx x ------------ = tan --+ C 1 + cos x 2

∫ dx x ------------ = - cot -- + C 1 - cos x 2

4. Prove that

∫ dx - 1 ---------= tanh x + C 1 - x2

where tanh x = sinhx-
coshx

. You may wish to show that 1 - tanh ²x = --1---
cosh2x

ﬁrst.

5. Consider the function f(x) for x a real positive number, deﬁned by the identity

∫ ∞ - t x f (x) = 0 e t dt

Use the techniques of this section to prove that

f (x) = x f (x), f (0) = 1

6. Use integration by parts to evaluate the integrals

∫ ∫ ---d---- ln x dx, x ln x

7. Evaluate the integral

∫ 2π -----dx------- 0 , ∣a ∣ > ∣b∣ a + b cos x

3 Line Integrals

Line integrals are used in physics 201 to compute work done by forces. The following is an exerpt from the physics 201 lecture note set.

The dynamical equations relating cause (forces) to effect (acceleration) in particle dynamics are technically second order differential equations

ma = F

2 d r(t) m -----2-- = F(r) dt

and in general second order differential equations are much harder to solve than ﬁrst order equations. The process of transforming this second order equation into a ﬁrst order one involves a single integration, which gives rise to a constant of integration called the energy.
Consider

ma = F

Now dot the velocity vector into both sides

ma ⋅ v = F ⋅ v

and use the identity

-d- -d- 1- a ⋅ v = ( v) ⋅ v = ( v ⋅ v) dt dt 2

We have manipulated Newton’s law into

d 1 dr ---( --mv ⋅ v) = mv ⋅ a = F ⋅ v = F ⋅ ---- dt 2 dt

This can be integrated

∫ tf d 1 ∫ tf dr ---(--mv ⋅ v) dt = F ⋅ ----dt t0 dt 2 t0 dt

let

dr dr r = r(t ), r = r(t ), v = (----)(t ), v = (----)(t ) 0 0 f f 0 dt 0 f dt f

be the initial and ﬁnal positions (and velocities) of the body as it is displaced along some path, while being acted upon by the force F, then

∫ ∫ tf dr-- rf t0 F ⋅ dt dt = r0 F ⋅ dr

and we arrive at the Work-Energy theorem;

∫ t d 1 1 1 ∫ r f ---(--mv ⋅ v) dt = --mv(tf ) ⋅ v(tf ) - --mv(t0) ⋅ v(t0) = f F ⋅ dr t0 dt 2 2 2 r0

The left hand side is called the change in kinetic energy, the right is called the work done by the force F If the force is conservative, we can apply the chain rule of calculus to the integral on the right-side of the work-energy theorem. A force is conservative if there exists a function called the potential V (r) such that application of the chain rule results in

d-- ∂V---dx-- ∂V--dy-- ∂V---dz-- - V (x, y, z) = - - - dt ∂x dt ∂y dt ∂z dt

dx dy dz = Fx ----+ Fy ----+ Fz ---- = F ⋅ v dt dt dt

This will mean that the components of F can be gotten directly from V ;

∂V--- ∂V--- ∂V--- F = (- , - , - ) = - ∇V ∂x ∂y ∂z

This may not be possible to do for a given force. Friction is a good example of a non-conservative force; there is no such function V (r) because technically the force of friction depends on both position r and velocity v of an object.

3.1 When does a force not have a potential?

If one cannot ﬁnd a function V such that

∂V ∂V Fx = - -----, Fy = - ----- ∂x ∂y

then the force is not conservative and has no potential. A good example is

F = - yi + xj = (- y, x)

We attempt to solve

∂V--- ∫ Fx = - y = - , - V = - y dx = - yx + C (y) ∂x

(since we integrate with respect to x, our “constant” of integration could depend on anything but x, including y). This potential does not give the correct F_y;

∂V--- -∂-- ′ Fy = x ⁄= - = - (yx - C (y)) = - x + C ∂y ∂x

On the other hand the force

F = yi + xj = (y, x)

is conservative and does have a potential function;

∫ ∂V--- Fx = y = - , - V = y dx = yx + C (y) ∂x

checking that this gives the correct F_y;

∂V ∂ F = x = - ----- = - ----( - yx - C (y)) = x + C ′ y ∂y ∂x

which works ﬁne of C(y) = C so C′ = 0. The potential that correctly gives both force components is

V (x, y) = - xy + C

Example

Find the work done by

F = yi + xj = (y, x)

when we move a mass from (0, 0) to (a,b) along the curve

x y = b( --)2 a

; Solution

W = - (V (a, b) - V (0, 0)) = - ((- ab + C ) - (- 0 + C )) = ab

(The work done against the force by the mover of the object is the opposite of this).

Example

Find the work done by

F = - yi + xj = (- y, x)

when we move a mass from (0, 0) to (a,b) along the curve

x y = b( --)2 a

; Solution We have no choice but to integrate. Parameterize the curve

2 x(t) = a t, y(t) = b t , 0 ≤ t ≤ 1

Find the velocity

dx-- vx(t) = = a, vy(t) = 2bt dt

ﬁnd the force at any point on the curve

Fx = - y = - bt2, Fy = x = at

and put it together; t₀ = 0, t_f = 1;

∫ ∫ 1 2 1 2 ab-- W = 0 ( - bt , at) ⋅ (a, 2bt) dt = 0 abt dt = 3

3.2 Problems

8.
A. For the force

a x a y F = --------------i + -------------j 1 (x2 + y2)2 (x2 + y2)2

determine the potential function (if there is one) and the work done in moving from (0, 1) to (1, 0) along the curve x² + y² = 1 (let a = 5.0).
B. For the force

a y a x F2 = --------------i - -------------j (x2 + y2)2 (x2 + y2)2

determine the potential function (if there is one) and the work done in moving from (0, 1) to (1, 0) along the curve x² + y² = 1 (let a = 5.0).

9.
Consider a force

N-- N-- F = Fxi + Fyj = 8.0 y i - 8.0 x j m m

If this force is conservative, ﬁnd its potential and use it to compute the work done by the force in moving a body along the curve y = 2 - 2x² from (0, 2) to (1, 0). If the potential does not exist, ﬁnd the work by integration.

10.
Consider a force

N-- N-- F = Fxi + Fyj = 8.0 x i - 8.0 y j m m

11.
A test for the existance of V ; for a force in two dimensions, if

∂Fx--- ∂Fy--- = ∂y ∂x

then there is a V since if

2 ∂V ∂V ∂Fx ∂Fy ∂ V Fx = - -----, Fy = - ----- then ------ = ------ = - -------- ∂x ∂y ∂y ∂x ∂x ∂y

Apply this to all of the forces above and test for a potential.

12.
Find the work done by the force

N F = (- y2i + x2j) ----- m2

in moving a particle from (x,y) = (1 m, 0) to (0, 1 m) along a circular path of radius 1 m.

13.
Find the work done by the force

2 2 N F = (- x i + y j) ---2- m

in moving a particle from (x,y) = (1 m, 0) to (0, 1 m) along a circular path of radius 1 m.

4 Examples of multiple integrals in 201

You will only see a certain type of multiple integrals, those that factor into products of integrals over different variables. This is nothing new, but one must be prepared for the shock of it.

Example

Compute the center of mass of a sheet of plywood of dimensions a by b with one corner at the origin.

Let the plywood have a mass per unit area ρ. Then a differential area of it has mass dm = ρdA. Break the board up into small elements, located at (x,y) of width dx and height dy. Add up all of the contributions