HP 33S Integral Demonstration and Programs

HP 33S Integral Demonstration and Programs

Integration

I'm going to start with a demonstration on how to calculate integrals with the HP 33S through the equation list.  The steps will be similar for the HP 32SII and the later HP 35S.  

1.  Enter the lower limit, press [ ENTER ], the higher limit.

2.  Press [ |→ ] [ STO ] (EQN), enter or select an equation.  

3.  Press [ |→ ] [ e^x ] ( ∫ ), to calculate the integration.  You will be prompted for the variable to integrate.  

Example:  Fresnel Integral 

Select Radians angle mode.  

Fresnel Sine:  

Equation:  SIN(SQ(T)), variable to integrate:  T   

Note:  [ x^2 ] -> SQ (square)

Use lower limit as 0.

S(2.4):   lower = 0, upper = 2.4,  S(2.4) ≈ 0.457864

S(5.8):   lower = 0, upper = 5.8,  S(5.8) ≈ 0.678004

Fresnel Cosine:

Equation:  COS(SQ(T)), variable to integrate:  T   

Use lower limit as 0.

C(2.4):   lower = 0, upper = 2.4,  C(2.4) ≈ 0.510157

C(5.8):   lower = 0, upper = 5.8,  C(5.8) ≈ 0.695845

Source:

"Fresnel Integral"  Wikipedia  https://en.wikipedia.org/wiki/Fresnel_integral  Retrieved May 2, 2022

Atmospheric Parameters

Given the height (H) in meters, the following can be estimated:

Temperature in Celsius  (°C):

T ≈ 15.04 - 0.00649 * H

Pressure in kilopascals (kPa):

P ≈ 101.29 * ((T + 273.15) / 288.08)^5.256

(note, the source had 273.1 but I had 27.3.15 for better accuracy)

Atmospheric Density (km/m^3):

D ≈ P/(0.2869 * (T + 273.15))

Program:

HP 33S:  Size: LN = 174, CK = 7625

A0001  LBL A

A0002  15.04

A0003  0.00649

A0004  INPUT H

A0005  ×

A0006  -

A0007  STO T

A0008  VIEW T

A0009  273.15

A0010  +

A0011  288.08

A0012  ÷

A0013  52.56

A0014  y^x

A0015 101.29

A0016  ×

A0017  STO P

A0018  VIEW P

A0019  0.2869

A0020  273.15

A0021  RCL+ T

A0022  ×

A0023  ÷

A0024  STO D

A0025  VIEW D

A0026  RTN

Example:

H = 50 m

Results:

T ≈ 14.715580 °C  (about 58.4879 °F)

P ≈ 100.894225 kPa

D ≈ 1.221648 kg/m^3

Source:

"Engineering Formula Sheet"  Project Lead The Way. https://www.madison-lake.k12.oh.us/userfiles/680/Classes/16192/IED-Review%20Engineering%20Formula%20Sheet.pdf   Last Retrieved April 29, 2022 

Legendre Polynomials

The value of Legendre Polynomials can be calculated using a closed formula from Rodrigues' formula:

P_n(x) = Σ( comb(n, k) * comb(n+k, k) * ((x - 1)/2)^k,  k= 0, n)

Program:  

HP 33S:  

LBL L:  Size:  LN = 30, CK = 14EC

LBL M: Size:  LN = 105, CK = AA05

Run XEQ L.  

L0001  LBL L

L0002  0

L0003  STO P

L0004  INPUT X

L0005  INPUT N

L0006  STO K

M0001  LBL M

M0002  RCL N

M0003  RCL K

M0004  nCr

M0005  RCL N

M0006  RCL+ K

M0007  LASTx

M0008  nCr

M0009  ×

M0010  -1

M0011  RCL+ X

M0012  2

M0013  ÷

M0014  RCL K

M0015  y^x

M0016  ×

M0017  STO+ P

M0018  DSE K

M0019  GTO M

M0020  1

M0021  STO+ P

M0022  RCL P

M0023  RTN

Examples:

N = 2, X = 0.25;  Result:  -0.406250

N = 3, X = -0.46;  Result:  0.446660

N = 4, X = 0.73;  Result:  -0.380952

Source:

"Legendre polynomials"  Wikipedia.  https://en.wikipedia.org/wiki/Legendre_polynomials  Updated April 6, 2022.  Last Accessed April 29, 2022.  

Eddie 

All original content copyright, © 2011-2022.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

March Calculus Madness Sweet Sixteen - Day 8: ∫ ln^2 x dx and ∫ ln^3 x dx

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Welcome to March Calculus Madness!

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Integration by parts to the rescue!

∫ ln^2 x dx

u = ln^2 x,   dv = dx

du = 2 * ln * 1/x dx,   v = x

∫ ln^2 x dx

= x * ln^2 x - ∫ 2 * ln x * 1/x * x dx

= x * ln^2 x - ∫ 2 * ln x dx

u = ln x, dv = 2 dx

du = 1/x dx, v = 2 * x

= x * ln^2 x - (2 * x * ln x - ∫ 1/x * 2 * x dx)

= x * ln^2 x - (2 * x * ln x - ∫ 2 dx)

= x * ln^2 x - 2 * x * ln x + ∫ 2 dx

= x * ln^2 x - 2 * x * ln x + 2 * x + C

∫ ln^3 x dx

u = ln^3 x dx, dv = dx

du = 3 * ln^2 x dx, v = x

= x * ln^3 x - ∫3 * ln^2 x * 1/x * x dx

= x * ln^3 x - 3 * ∫ ln^2 x dx

= x * ln^3 x - 3 * (x * ln^2 x - 2 * x * ln x + 2 * x + C)

(see the above)

=  x * ln^3 x - 3 * x * ln^2 x + 6 * x * ln x - 6 * x + D   

(where D = 3 * C, C and D are constants)

Eddie

All original content copyright, © 2011-2022.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

March Calculus Madness Sweet Sixteen - Day 7: ∫ 1/(1 + cos x) dx and ∫ 1/(1 + sin x) dx

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Welcome to March Calculus Madness!

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∫ 1/(1 + cos x) dx and ∫ 1/(1 + sin x) dx

To tackle these integrals, we will make use of the following derivatives and trig identities:

d/dx tan x = sec^2 x dx

d/dx cot x = -csc^2 x dx

d/dx csc x = -cot x * csc x

d/dx sec x = tan x * sec x

sin^2 x + cos^2 x = 1

cot x = 1/tan x,  csc x = 1/sin x,  sec x = 1/cos x

∫ 1/(1 + cos x) dx

= ∫ 1/(1 + cos x) * (1 - cos x)/(1 - cos x) dx

= ∫ (1 - cos x)/(1 - cos^2 x) dx

= ∫ (1 - cos x)/sin^2 x dx

= ∫ csc^2 x - cot x * csc x dx

= -cot x + csc x + C

∫ 1/(1 + sin x) dx

= ∫ 1/(1 + sin x) * (1 - sin x)/(1 - sin^2 x) dx

= ∫ (1 - sin x)/(1 - sin^2 x) dx

= ∫ (1 - sin x)/(cos^2 x) dx

= ∫ sec^2 x - tan x * sec x dx

= tan x - sec x + C

Note:  CAS systems in graphing calculators will default to using sin x, cos x, and tan x.  

Eddie

All original content copyright, © 2011-2022.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

March Calculus Madness Sweet Sixteen - Day 6: ∫ √a ÷ (√(x+a) - √x) dx

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Welcome to March Calculus Madness!

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∫ √a ÷ (√(x+a) - √x) dx

Simplify:

√a ÷ (√(x+a) - √x)

Multiply by (√(x+a) + √x) ÷ (√(x+a) + √x) 

√a ÷ (√(x+a) - √x)

= √a ÷ (√(x+a) - √x) * (√(x+a) + √x) ÷ (√(x+a) + √x) 

= √a * (√(x+a) + √x) ÷ (x + a - x)

= 1 ÷ √a * (√(x+a) + √x)

Then:

∫ √a ÷ (√(x+a) + √x) dx

=  ∫ 1 ÷ √a * (√(x+a) + √x) dx

= a^(-1/2) * ∫ (x+a)^(1/2) + x^(1/2) dx

= 2 ÷ (3 * a^(1/2)) * ((x + a)^(3/2) + x^(3/2)) + C

= 2 ÷ (3 * a^0.5) * ((x + a)^1.5 + x^1.5) + C

Questions of preference:

Which is your preferred notation:

(1)  √x,  x^(1/2), or x^0.5?

(2)  x^(3/2) or x^1.5?  

Let us know in the comments.

Eddie 

All original content copyright, © 2011-2022.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

Working with Laplace Transforms

Working with Laplace Transforms


Using Laplace Transforms to solve various differential equations


Table of Laplace Transforms

Function:  y(t)
Transformed Function:  f(s)

f(s) = ℒ( y(t) )
= ∫ y(t) ∙ exp(-s ∙ t) dt from t = 0 to t = ∞,  s is a complex number

a1, a2, and a are constants

ℒ( a1 ∙ y1(t) + a2 ∙ y2(t) ) = a1 ∙ f1(s) + a2 ∙f2(s)

ℒ ( y'(t) ) = s ∙ f(s) - y(0)

ℒ( t ∙ y(t) ) = -f'(s)

ℒ( 1 ) = 1 ÷ s

ℒ( a ) = a ÷ s

ℒ( t^n ) = n! ÷ (s^(n+1))

ℒ( t^n ÷ n! ) = 1 ÷ (s^(n+1))

ℒ( e^(a ∙ t) ) = 1 ÷ (s - a)

ℒ( sin(a ∙ t) ) = a ÷ (s^2 + a^2)

ℒ( cos(a ∙ t) ) = s ÷ (s^2 + a^2)


All the problems presented today will have the initial condition y(0) = 0.  



Problem 1:  y'(t) + a ∙ y(t) = b with y(0) = 0

Note: y'(t) = dy/dt.   All of the problems this weekend will have an initial condition.

 y'(t) + a ∙ y(t) = b

Applying the Laplace Transform:

s ∙ f(s) - y(0) + a ∙ f(s) = b ÷ s

f(s) ∙ (s + a) = b ÷ s

f(s) = b ÷ (s ∙ (s + a))



Partial fractions:

b ÷ (s ∙ (s + a)) = G ÷ s + H ÷ (s + a)

Simplifying and considering the numerator:

b + 0 ∙ s = a ∙ G + s ∙ (G + H)

Then:

b = a ∙ G
0 = G + H

G = b/a,  H = -G, H = -b/a


f(s) = (b/a) ÷ s - (b/a) ÷ (s + a)

Reverse the Laplace Transform:

y(t) = b/a - b/a ∙ e^(a ∙ t)

y(t) = b/a ∙ (1 - e^(a ∙ t))



Problem 2:  y'(t) + a ∙ y(t) = e^(b ∙ t) with y(0) = 0

y'(t) + a ∙ y(t) = e^(b ∙ t)

Applying the Laplace Transform:

s ∙ (f(s) - y(0) + a ∙ f(s) = 1 ÷ (s - b)

(s + a) ∙ f(s) = 1 ÷ (s - b)

f(s) = 1 ÷ ((s - b) ∙ (s + a))


Partial fractions:

1 ÷ ((s - b) ∙ (s + a)) = E ÷ (s - a) + F ÷ (s + a)

Simplifying and considering the numerator:

1 = E ∙ (s + a) + F ∙ (s - b)

0 ∙ t + 1 = E ∙ (s + a) + F ∙ (s - b)

Solving for E and F:

1 = a ∙ E - b ∙ F
0 = E + F

E = -F;  F = -1 ÷ (a + b), E = 1 ÷ (a + b)


f(s) = (1/(a + b)) ÷ (s - b) - (1/(a - b)) ÷ (s + a)

Reverse the Laplace Transform:

y(t) = e^(b ∙ t) ÷ (a + b) - e^(-a ∙ t) ÷ (a + b)

y(t) = (e^(b ∙ t) - e^(-a ∙ t)) ÷ (a + b)



Problem 3:  y'(t) + a ∙ y(t) = b ∙ t + c with y(0) = 0

y'(t) + a ∙ y(t) = b ∙ t + c

Applying the Laplace Transform:

s ∙ f(s) - y(0) + a ∙ f(s) = (b ∙ 1!) ÷ (s^(1 + 1)) + c ÷ s

s ∙ f(s) + a ∙ f(s) = b ÷ s^2 + c ÷ s

f(s) = (b + c ∙ s) ÷ (s^2 ∙ (s + a))


Partial fractions:

(b + c ∙ s) ÷ (s^2 ∙ (s + a)) = G ÷ (s + a) + (s ∙ H + J) ÷ s^2

= G ÷ (s + a) + H ÷ s + J ÷ s^2

Simplifying and considering the numerator:
 
b + c ∙ s + 0 ∙ s^2  = G ∙ s^2 + H ∙ s^2 + H ∙ a ∙ s + J ∙ s + J ∙ a

b = a ∙ J
c = a ∙ H + J
0 = G + H

H = -G; J = b/a, H = (a ∙ c - b) ÷ a^2, G = (b - a ∙ c) ÷ a^2


f(s) = ((b - a ∙ c) ÷ a^2) ÷ (s + a) + ((a ∙ c - b) ÷ a^2) ÷ s + (b/a) ÷ s^2

Reverse the Laplace Transform:

y(t) = ((b - a ∙ c) ÷ a^2) ∙ e^(-a ∙ t) + ((a ∙ c - b) ÷ a^2)  + (b/a) ∙ t





Problem 4:  y'(t) + a ∙ y(t) = b ∙ t^2 with y(0) = 0

y'(t) + a ∙ y(t) = b ∙ t^2

Applying the Laplace Transform:

s ∙ f(s) - y(0) + a ∙ f(s) = 2 ∙ b ÷ s^3

f(s) = (2 ∙ b) ÷ (s^3 ∙ (s + a))


Partial fractions:

(2 ∙ b) ÷ (s^3 ∙ (s + a)) = (F ∙ s^2 + G ∙ s + H) ÷ s^3 + K ÷ (s + a)

(2 ∙ b) ÷ (s^3 ∙ (s + a)) = F ÷ s + G ÷ s^2 + H ÷ s^3 + K ÷ (s + a)

Simplifying and considering the numerator:

0 ∙ s^3 + 0 ∙ s^2 + 0 ∙ s + 2 ∙ b = F ∙ s^3 + G ∙ s^2 + H ∙ s + F ∙ a ∙ s^2 + G ∙ a ∙ s + H ∙ a + K ∙ s^3

2 ∙ b = a ∙ H
0 = H + a ∙ G
0 = G + a ∙ F
0 = F + K

F = -K; K = (-2 ∙ b) ÷ a^3 ; F = (2 ∙ b) ÷ a^3; H = (2 ∙ b) ÷ a; G = (-2 ∙ b) ÷ a


f(s) = (2 ∙ b)/a^3 ÷ s - (2 ∙ b)/a^2 ÷ s^2 + (2 ∙ b)/a ÷ s^3 - (2 ∙ b)/a^3 ÷ (s + a)

I use Wolfram Alpha and CAS calculators to help verify whether the work is done correctly.  Sometimes I have deal with problems in parts.  

Reverse the Laplace Transform:

y(t) = (2 ∙ b)/a^3 - (2 ∙ b)/a^2 ∙ t + (b/a) ∙ t^2 - (2 ∙ b)/a^3 ∙ e^(-a ∙ t)

Note:  ℒ( t^n ÷ n! ) = 1 ÷ (s^(n+1)), ℒ( t^2 ÷ 2) = 1 ÷ (s^2)


Eddie



All original content copyright, © 2011-2022.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.

Six Integrals from Calculus I: Substitution vs Integration by Parts

Six Integrals from Calculus I:  Substitution vs Integration by Parts

Even slight differences integrands can change the strategy.  Observe.

Substitution

For these three integrals, the substitution u = x^2 will serve well.

Observe:

u = x^2

du = 2 x dx

du / 2 = x dx

∫ x sin x^2 dx

= ∫ 1/2 sin u du

= -1/2 cos u + C

= -1/2 cos x^2 + C

∫ x cos x^2 dx

= ∫ 1/2 cos u du

= 1/2 sin u + C

= 1/2 sin x^2 + C

∫ x e^(x^2) dx

= ∫ 1/2 e^u du

= 1/2 e^u + C

= 1/2 e^(x^2) + C

Integration by Parts

General:   ∫ u dv = u v - ∫ v du,  v and u are functions (usually of x)

∫ x^2 sin x dx

u = x^2, dv = sin x dx;   du = 2 x dx, v = - cos x

= -x^2 cos x + ∫ 2 x cos x dx

u = 2 x, dv = -cos x dx;  du = 2 dx, v = -sin x

= -x^2 cos x + 2 x sin x - ∫ 2 sin x dx

= -x^2 cos x + 2 x sin x - ∫ 2 sin x dx

∫ x^2 cos x dx

u = x^2, dv = cos x dx;   du = 2 x dx, v = sin x

= x^2 sin x - ∫ 2 x sin x dx

u = 2 x, dv = sin x dx;  du = 2 dx, v = -cos x

= x^2 sin x + 2 x cos x - ∫ 2 cos x dx

= x^2 sin x + 2 x cos x - 2 sin x + C

∫ x^2 e^x dx

u = x^2, dv = e^x dx;  du = 2 x dx, v = e^x

= x^2 e^x - ∫2 x e^x dx

u = 2 x, dv = e^x dx;  du = 2 dx, v = e^x

= x^2 e^x - 2 x e^x + ∫2 e^x dx

= x^2 e^x - 2 x e^x + 2 e^x + C

= e^x (x^2 - 2 x + 2) + C

Bottom line, being good at integration, like everywhere else in math, will require practice, experience, and maybe later in life, some retraining.  

Students, best of luck in your studies, in class and in life.  

Eddie 

All original content copyright, © 2011-2021.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

Retro Review: Texas Instruments TI-36X II

Retro Review:  Texas Instruments TI-36X II 

Quick Facts:

Model:  TI-36 XII

Company:  Texas Instruments

Years:  1999 - mid-2000s

Battery:  CR2025

Operating System:  Algebraic

Number of Memory Registers:  5: A, B, C, D, E

Datamath Calculator Museum page:  http://www.datamath.org/Sci/Modern/TI-36X-II.htm

Keyboard

I like the dark gray keyboard, giving a calculator a sleek look.  The dark mint green and white font on the keys and the keyboard provides a nice contrast.  The keys are a pleasure to push, however make sure the key registers on the screen before the next key is pushed; there is no key buffer.  

To keep the keyboard from possibly being cluttered, most of the funcitons are stored in menus.  This includes:

LOG:  log, 10^; ln, e^

HYP:  sinh, sinh^-1; cosh, cosh^-1; tanh, tanh^-1

COMPX (complex numbers):  conj (conjugate), real (real part); imag (imaginary part), abs (absolute value)

TRIG: sin, sin^-1; cos, cos^-1; tan, tan^-1

LOGIC:  and, or, xor; not, 2's

PRB (probability):  nPr (permutations), nCr (combinations), ! (factorial of positive integers up to 69); rand (random numbers between 0 to 1), randi (random integers between a and b)

Features

Base Conversion with Boolean Functions:  The TI-36X II overs conversions between integers in decimal (base 10), hexadecimal (base 16), or octal (base 8).  Somehow binary isn't included.  We get the nice basic set of Boolean functions.  The hexadecimal integers A through F are located on the shifted functions of keys [ 1 ] through [ 6 ].

Fractions:  Use the [A b/c] key to enter fractions.  There are two conversion functions:  F<>D converts between fractions and decimals, while A b/c<>d/e converts between mixed form and improper form.   Fractions are simplified automatically (example: entering 12/24 returns 1/2).

Operation Storage:  This is the programmable feature of the TI-36X II.  Two operations, OP1 and OP2, allow the user to store a simple operation (+, -, *, /, ^) followed by a number.  Operations are stored with >OP1 and >OP2.  

Example:  If I store +5 to OP1 and *2 to OP2, then: 

7 [ OP1 ] returns 12

7 [ OP2 ] returns 14

7 [ OP1 ] [ OP2 ] returns 24  (7+5=12, 12*2=24)

7 [ OP2 ] [ OP1 ] returns 19  (7*2=14, 14+5=19)

Conversions:  The TI-36X II features 20 metric-U.S. conversions.  It is really interesting how they are paired up.   Each conversion set is represented once, like cm←→in.  Pressing [ = ] gives the conversion "left to right", like cm to in.  However, pressing [ 2nd ] [ = ] gives the conversion "right to left", in to cm.

Constants:  The TI-36X II has 16 conversions in SI units.  

Complex Numbers:  Complex numbers are typed as ordered pairs:  (real, imaginary).   The complex operations are limited to arithmetic (+, -, x , /) and the COMPX menu only.  Strangely, there is no arg/angle function.

Integrals:  Integrals are done in Radian angle mode.   The variable of integration is A.   The steps to perform an integration:

1.  Store the lower limit in the variable A.

2.  Store the upper limit in the variable B.

3.  Store the number of divisions in the variable C.  The Simpson's rule is used, so the number of divisions is even, and the more divisions are used, the more accurate the answer but the longer the calculation takes.    

4.  Press the [ ∫dx ] key.  Type the function (in terms of A, use [MEMVAR], A).  Press [ = ].  During calculation, the calculator shows a clock icon with "CALC" displayed. 

The function must be continuous and defined everywhere.  

Example:  ∫ 4A^3 - 5e^A dA, lower (A) = 1, upper (C) = 5

C = 10 calculates the integral as -104.480831

C = 20 calculates the integral as -104.4747906

After each integral, the C is cleared  and the value of A matches value the B.  

Rectangular/Polar Conversions:  Like higher-end Texas Instrument calculators, conversions are handled on four functions:

R>Px(x, y)

R>Py(x, y)

P>Rx(r, θ)

P>Ry(r, θ)

Statistics:  The TI-36X II has four regression models:

Linear (LIN):  y = a + bx

Logarithmic (LN):  y = a + b ln x

Exponential (EXP):  y = a * b^x

Power (PWR):  y = a*x^b

A model, 1-Variable (1-VAR), or 2-Variable (2-VAR) must be selected first before data can be entered.  Up to 42 data points can be entered.  

Verdict

The TI-36X II is a predecessor to the TI-36X Pro and is a great calculator to use.  There are some missing functions that I think should have been present were the argument (angle) function, the conversion to binary integers, maybe integer and fraction parts.   I am not a fan of selecting variables through the menu, I personally prefer the use of ALPHA key.

Still a great calculator for the collection.  $20, $25 at most should be a good price, even though it is one of the rarer calculators from Texas Instruments. 

 

Eddie

All original content copyright, © 2011-2021.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

Casio fx-9750GIII: Trigonometric Integral Graphs, including The Euler Spiral

 Casio fx-9750GIII:  Trigonometric Integral Graphs, including The Euler Spiral

Introduction

The program ESGRAPH plots the parametric equations containing integrals:

x(t) = ∫(cos f(x) dx, x = 0 to t)

y(t) = ∫(sin f(x) dx, x = 0 to t)

The program uses f(x) = a*x^p

When p = 2, the curve is a Euler Spiral (see source).

Casio fx-9750GIII Program: ESGRAPH

' 2020-12-22 EWS

Rad

' L1, X L2, Y

"F = A×X^P"

"FACTOR"?→A

"P=2, EULER SPIRAL"

"POWER"?→P

∫(A×cos(X^P),0,-2π)→X

∫(A×sin(X^P),0,-2π)→Y

{X}→List 1

{Y}→List 2

Menu "π÷R?","8",A,"16",B,"32",C

Lbl A: 8 → R: Goto 0

Lbl B: 16 → R: Goto 0

Lbl C: 32 → R : Goto 0

Lbl 0

For -2π+π÷R→T To 2π Step π÷R

∫(A×cos(X^P),0,-2π)→X

∫(A×sin(X^P),0,-2π)→Y

Augment(List 1, {X})→List 1

Augment(List 2, {Y})→List 2

(T+2π)÷(4π)×100→G

RndFix(G,0)→G

Locate 1,7,"PROGRESS:    %"

Locate 11,7,G

Next

S-Gph1 DrawOn, xyLine, List 1, List 2, 1, Dot

DrawStat

Notes:

*  There are four spaces between PROGRESS: and %.  This progress meter is to inform the user of the program's completion. 

*  The characters % and ', percent sign and apostrophe respectively, are found on the program editing default menu (TOP, BTM, SRC, MENU, A ←→a, CHAR), by pressing [F6] (CHAR), [F2] (SYBL)

*  Everything that follows an apostrophe is a comment and can be left out.

*  The range of -2π to 2π is used. 

*  The lower the precision, the faster the graph is made but accuracy of the graph is sacrificed.  In a lot of cases, patience is required. 

*  P is a positive integer.  

Examples

Example 1:  Euler Spiral:  A = 1/2, P = 2, precision 32

Example 2:  A = 1, P  = 3, precision 32

Example 3:  A = 2, P = 1, precision 16

Source:

Havil, Julian.  Curves for the Mathematical Curious: An Anthology of the Unpredictable, Historical, Beautiful, and Romantic  Princeton Universal Press. Princeton and Oxford. Princeton, NJ 2019 ISBN 978-0-691-18005-2

Eddie

All original content copyright, © 2011-2021.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

The Error Function and Normal Distribution: Norm(x) and Erf(x)

The Error Function and Normal Distribution:  Norm(x) and Erf(x)

Introduction

The error function is defined as:

erf(x) = 2/√π * ∫( e^(-t^2) dt, 0, x)

norm(x) = 1/√(2*π) * ∫( e^(-z^2/2) dx, -∞, x)

We were assuming that the mean μ = 0 and variance σ = 1.

Derivation:  norm(x) in terms of erf(x)

norm(x)

= 1/√(2*π) * ∫( e^(-z^2/2) dx, -∞, x)

= 1/√(2*π) * ∫( e^(-z^2/2) dx, -∞, 0) + 1/√(2*π) * ∫( e^(-z^2/2) dx, 0, x)

= 1/2 + 1/√(2*π) * ∫( e^(-z^2/2) dx, 0, x)

-----------------------------
Substitution:

t^2 = z^2/2
t = z/√2
√2 * t = z
√2 dt = dz

z = 0,  t = 0
z = x,  t = x/√2
-----------------------------

= 1/2 + 1/√(2*π) * ∫( e^(-z^2/2) dx, 0, x)

= 1/2 + 1/√(2*π) * ∫(√2 *  e^(-t^2) dt, 0, x/√2)

 = 1/2 + 1/√π * √π/2 * 2/√π * ∫(√2 *  e^(-t^2) dt, 0, x/√2)

 = 1/2 + (1/√π * √π/2) * (2/√π * ∫(√2 *  e^(-t^2) dt, 0, x/√2))

 = 1/2 + 1/2 * 2/√π * ∫(√2 *  e^(-t^2) dt, 0, x/√2)

= 1/2 + 1/2 *erf(x/√2)

norm(x) = 1/2 + 1/2 *erf(x/√2)

norm(x) - 1/2 = 1/2 *erf(x/√2)

2 norm(x) - 1 = erf(x/√2)

Let t = x/√2, then:

2 norm(√2 * t) - 1 = erf(t)

Summary

norm(x) = 1/2 + 1/2 * erf(x/√2)

erf(t) = 2 * norm(√2 * t) - 1 

with μ = 0 and σ = 1.

Examples

norm(1) = 1/2 + 1/2 * erf(1/√2) ≈ 0.841344746069

erf(1) = 2 * norm(√2) - 1 ≈ 0.84270079295


Eddie

All original content copyright, © 2011-2019.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.

Integrals with the Error (erf) Function

Integrals with the Error (erf) Function

Introduction - Integrals Involving the Erf Function

erf(x) = 2 / √π * ∫( e^-(t^2), t, 0, x)   ** graphing calculator syntax


Problem 1:  ∫( e^-(t^2), t, a, x) with a > 0 and x > a

∫( e^-(t^2), t, 0, x) = ∫( e^-(t^2), t, 0, a)  + ∫( e^-(t^2), t, a, x)

 ∫( e^-(t^2), t, 0, x) - ∫( e^-(t^2), t, 0, a)  = ∫( e^-(t^2), t, a, x)

√π/2 * 2/√π * (  ∫( e^-(t^2), t, 0, x) - ∫( e^-(t^2), t, 0, a) ) =   √π/2 * 2/√π * ∫( e^-(t^2), t, a, x)

√π/2 * ( 2/√π *  ∫( e^-(t^2), t, 0, x) - 2/√π *  ∫( e^-(t^2), t, 0, a) ) =   ∫( e^-(t^2), t, a, x)

√π/2 * ( erf(x) - erf(a)  ) =   ∫( e^-(t^2), t, a, x)


 ∫( e^-(t^2), t, a, x) = √π/2 * (erf(x) - erf(a))

Problem 2:  ∫( e^-((t + b)^2), t, 0, x)

∫( e^-((t + b)^2), t, 0, x)

Substitutions:
u = t + b
u - b = t
du = dt

Limits:
t  = x, u = x + b
t = 0, u = b

∫( e^-((t + b)^2), t, 0, x)

= ∫( e^-(u^2), u, b, b+x)

Note √π/2  * 2 /√π  = 1

=  √π/2  * 2 /√π  * ∫( e^-(u^2), u, b, b+x)

Using problem 1 as an aid to help us get:

= √π/2  * (eft(x + b) - erf(x))

∫( e^-((t + b)^2), t, 0, x) = √π/2  * (eft(x + b) - erf(x))

Problem 3:   ∫( e^(-a*t^2 + b), t, 0, x)

∫( e^(-a*t^2 + b), t, 0, x)

= ∫( e^(-a*t^2) * e^b, t, 0, x)

e^b is a constant.

= e^b * ∫( e^(-a*t^2), t, 0, x)

Substitutions:
u = √a * t
u^2 = a * t^2

du = dt * √a
du / √a = dt

Limits:
t  = x, u = √a * x
t = 0, u = 0

= e^b * 1/√a  * ∫( e^(-u^2), u, 0, √a * x)

= e^b * 1/√a  * √π/2  * 2 /√π  * ∫( e^(-u^2), u, 0, √a * x)

= e^b * 1/√a  * √π/2  * erf(√a * x)

∫( e^(-a*t^2 + b), t, 0, x) = e^b * 1/√a  * √π/2  * erf(√a * x)

Problem 4:  ∫( e^-(ln^2 t)/t, t, 1, x)

∫( e^-(ln^2 t)/t, t, 1, x)

Substitutions:
u^2 = ln^2 t
u = ln t
du = 1/t dt
t  du = dt

1/t * t = 1

Limits:
 t = x, u = ln x
t = 1, u = ln 1 = 0

= ∫( e^-(u^2), u, 0, ln x)

= √π/2  * 2 /√π  * ∫( e^-(u^2), u, 0, ln x)

= √π/2  * erf(ln x)


∫( e^-(ln^2 t)/t, t, 1, x) = √π/2  * erf(ln x)

Problem 5:  2/√π * ∫(e^(t^2), t, 0, x)

2/√π * ∫(e^(t^2), t, 0, x)

Substitutions:
-u^2 = t^2
u^2 = -t^2
u = i * t   where i = √ -1
du = i dt
- i du = dt  because 1/i = -i

Limits:
t = x, u = i *x
t = 0, u = 0

= 2/√π * ∫(e^-(u^2) * -i, t, 0, i*x)

= 2/√π * -i * ∫(e^-(u^2), t, 0, i*x)

= -i * erf(i*x)

2/√π * ∫(e^(t^2), t, 0, x) = -i * erf(i * x)

Alternative substitution:
-u^2 = t^2
i * u = t
u = -i * t
du = -i dt
i du = dt

Limits:
t = x, u = -i *x
t = 0, u = 0

Then we get

2/√π * ∫(e^(t^2), t, 0, x)

= 2/√π * i * ∫(e^-(u^2), t, 0,- i*x)

= i * erf(-i * x)

In summary....

2/√π * ∫(e^(t^2), t, 0, x) = -i * erf(i * x) = i * erf(-i * x)

These are several of integrals using the erf function. - Eddie

All original content copyright, © 2011-2019.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.

HP Prime: Firmware Update (13865, date 2018.07.06)

HP Prime:  Firmware Update (13865, date 2018.07.06)

There is a new update to the HP Prime firmware.  The new version is 13865. 

You can find the file and details here:

Details and file: 

http://www.hpmuseum.org/forum/thread-11097.html

http://www.hpmuseum.org/forum/thread-11092.html

Firmware download:  ftp://ftp.hp.com/pub/calculators/Prime/

I haven’t had much chance to work with the new firmware, however according to the release information, some of the highlights are:

* There is a new red indicator when the HP Prime’s power reaches below 10%.

* All integrals in HOME mode will calculate numerical results.  Previously results are returned in either numerical or exact answers.  Exact answers will still be provided in CAS mode.

* You can control the amount of time until the HP Prime dims its screen.  The screen is dimmed when you do nothing on the calculator after a set amount of time.  The time is set in milliseconds as a base integer (30,000 in decimal or 7530_16 in hexadecimal).  The variable is TDim found in the Vars-Setting menu. 

* The command EVAL is said to now help INPUT with local variables.  I haven’t played around with this yet. 

* Updated CAS and improvements

As a note, there the HP Prime now gets a new processor G2.   To find out what hardware version you have, press [Help], press the soft key (Tree), scroll up, select About HP Prime.  I have hardware revision C.  The threads on the MoHPC site goes into more details.  The reason why I mention this is if you have trouble updating your HP Prime, the firmware files may need to be in a new folder.  The folder is:

Hardware C (and I’m assuming this will work for A):

(your drive) \HP Connectivity Kit\Firmware\PrimeG1

Thank you to HP and HP Museum of Calculators! 

Happy Computing,

Eddie

All original content copyright, © 2011-2018.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.  Please contact the author if you have questions.

Area Under the Curve Using Parametric Curves: Casio Classpad fx-CP400 and HP Prime (CAS)

Area Under the Curve Using Parametric Curves:   Casio Classpad fx-CP400 and HP Prime (CAS)

The CAS (computer algebra system) function paramint calculates the area under the curve of a function defined as the parametric curve (x(t), y(t)).  The integral is  ∫ y(t) * x’(t) dt .

Casio Classpad fx-CP400:  paramint

It would the most efficient way is use the Define command on the home screen (like the screen shot above). 

Define paramint(xt,yt,t0,t1) = ∫ (yt * diff(xt,t) dt, t = t0 to t = t1)

HP Prime:  paramint  (CAS Program)

The program paramint is going to be defined as CAS function.  When first creating paramint, check the CAS box before pressing the soft key (OK).  Also pay attention to the letters being typed while using the ALPHA key.  In the program editor, letters are defaulted to uppercase while CAS variables are lowercase.  Finally, commands called from the Toolbox in Program editing mode will have a “CAS.” suffix.  Erase the suffices. 

#cas

paramint(xt,yt,t0,t1):=

BEGIN

LOCAL n;

n:=int(yt*diff(xt,t),t,t0,t1);

return n;

END;

#end

Examples

Example 1:  x(t) = 2t^3 + 1, y(t) = t^2, t0 = 0, t1 = 1

Result:  6/5

Example 2:  x(t) = sin t, y(t) = 3 cos t, t0 = 0, t1 = π

Result: 3/2*π ≈ 4.71238898039

This blog is property of Edward Shore, 2016

HP RPN Advance Functions Video Up and Running

This is part two of the RPN Mode for HP Prime tutorial.  This time I am taking about advanced functions, such as numeric integrals, numeric derivatives, summation, and solving equations.

Link to the video:

http://youtu.be/YAJwjpldiak

DETAILS:


Syntax - RPN Mode:
-----
MAX and MIN
Path:  Toolbox, Math, 2, 1 for MAX (2 for MIN)
Syntax:
1:  list or vector
MAX(1) or MIN(1)
 - or -
Numbers on the stack
MAX(n) or MIN(n)   (n is 9 or less)
-----
Factorizing Integers – CAS.ifactor
1.  Enter the integer to be factorized, press the Enter key.
2.  Press Toolbox, CAS, 5, 2, 1 to execute CAS.ifactor(1)
-----
Generating a Sequence - MAKELIST
Path:  Toolbox, Math, 6, 1
Stack:
5:  ‘function in single quotes’
4:  ‘variable in single quotes’
3:  beginning point
2:  ending point
1:  increment/decrement
MAKELIST(5)
-----
Finding the Roots of a Polynomial - CAS. proot
Path:  Toolbox, CAS, 6, 1
Stack:
1:  vector or list of coefficients (decreasing power of x)
CAS.proot(1)
-----
Summation – CAS.sum
Path: Toolbox, Math, 2, 5
Stack:
4:  ‘function in single quotes’
3:  ‘variable’
2:  starting point
1:  finishing point
CAS.sum(4)
-----
Numerical Integration
1.  Press the Template Key, choose ∫ (2nd row, 4th column)
2.  Fill in the template – use capital letters for the variable
3.  Press the Enter Key
4.  Finally, press Toolbox, CAS, 1, 1 to execute CAS.simplify(1)
-----
Numerical Derivative
1.  Store value in the variable: Number, Enter Key, Variable in single Quotes, Shift Key, EEX Key
2.  Press the Template Key, choose ∂ (1st row, 4th column).  Fill in the template – use capital letters
for the variable, and press the Enter Key
3.  Finally, press Toolbox, CAS, 1, 1 to execute CAS.simplify(1)
-----
Matrix Operations
Put the matrix on the stack.  Use the Template Key, choose matrix (1st row, 6th column).
Determinant:  Press Toolbox, Math, 7, 2.  Determinant is calculated automatically. 
Inverse:  Press Shift, then the division key. (x^-1).  This works for square matrices only.
Eigenvalues:  Press Toolbox, Math, 7, 6, 1.  CAS.EIGENVAL is called.  Execute CAS.EIGENVAL(1).
-----
Finding Numerical Roots using FNROOT
Path:  Toolbox, Math, 2, 4
Stack:
3:  ‘function in single quotes’
2:  ‘variable’
1:  guess
Execute FNROOT(3)
Format:   function = 0
-----

Eddie


This blog is property of Edward Shore.  2014
 

Calculus Revisited #14: Integration - Decomposition of Fractions

Welcome to Part 14 of our 21 part of our Calculus Revisited. Today's blog entry involves integration using decomposition of fractions.

Decomposition of Fractions

Here is the general strategy.

If the denominator has (x - c)^r then:

α / (x - c)^r = A1 / (x - c) + A2 / (x - c)^2 + ... + Ar / (x - c)^r

If the denominator has (x^2 + cx + d)^r then:

β / (x^2 + cx + d) = (A1x + B1) / (x^2 + cx + d) + (A2 x + B2) / (x^2 + cx + d)^2 + ... + (Ar x + Br) / (x^2 + cx + d)^r

Here are some examples.

Problems

Sometimes polynomial division is all you need.

1. ∫ (8x - 3) / (4x + 1) dx

First execute the division (8x - 3) / (4x + 1).

(8x - 3) / (4x + 1) = 2 - 5 / (4x + 1)

So

∫ (8x - 3) / (4x + 1) dx
= ∫ 2 - 5 / (4x + 1) dx
= 2x - 5/4 * ln(4x + 1)

Final: ∫ (8x - 3) / (4x + 1) dx = 2x - 5/4 * ln(4x+ 1) + C


Now here are two basic examples of decomposition of fractions.

2. ∫ dx / ((x + 1)(2x + 3))

Then:

A / (x + 1) + B / (2x + 3 ) = 1 / ((x + 1)(2x + 3))

A (2x + 3) + B (x + 1) = 1

2A(x) + 3A + B(x) + B = 1 + 0(x)

Look at the coefficients and match them.

Constant: 3A + B = 1
Coef. of x: 2A + B = 0

The next task is to solve for A and B.

3A + B = 1
2A + B = 0

Subtract the second equation from the first equation

A = 1

Find B

Using the first equation:
3(1) + B = 1
B = 1 -3
B = -2

Using the second equation:
2(1) + B = 0
2 + B = 0
B = -2

Conclude A = 1, B = -2

Back to our integral:

∫ dx / ((x + 1)(2x + 3))
= ∫ A / (x + 1) + B / (2x + 3) dx
= ∫ 1 / (x + 1) - 2 / (2x + 3) dx
= ln(x + 1) - ln(2x + 3)
= ln( (x + 1) / (2x + 3) )

Final: ∫ dx / ((x + 1)(2x + 3)) = = ln( (x + 1) / (2x + 3) ) + C

3. ∫ (2x + 1) / ( (x^2 + 1) (x - 1) ) dx

Decomposing:

(2x + 1) / ( (x^2 + 1) (x - 1) ) = (Ax + B) / (x^2 + 1) + C / (x - 1)

Then:

(Ax + B) (x - 1) + C (x^2 + 1) = 2x + 1
A(x^2) - A(x) + B(x) - B + C(x^2) + C = 2x + 1 + 0(x^2)
(A + C)x^2 + (-A + B)x + (-B + C) = (0)x^2 + (2)x + 1

Matching up terms of x^2, x, and the constant:
A + C = 0
-A + B = 2
-B + C = 1


From the first equation, C = -A (alternatively A = -C)

Using the second and third equations:
-A + B = 2
-B + C = 1

-A + B = 2
-B + (-A) = 1

-A + B = 2
-A - B = 1

Add the two equations together to get

-2A = 3
A = -3/2

Then: -A - B = 1
-(-3/2) - B = 1
- B = 1 - 3/2
B = 3/2 - 1
B = 1/2

So A = -3/2, B = 1/2, and C = 3/2

And:
∫ (2x + 1) / ( (x^2 + 1) (x - 1) ) dx
= ∫ (Ax + B) / (x^2 + 1) + C / (x - 1) dx
= ∫ ((-3/2)x + 1/2) / (x^2 + 1) + (3/2) / (x - 1) dx
= ∫ (-3/2)x / (x^2 + 1) + (1/2) / (x^2 + 1) + (3/2) / (x - 1) dx
= (-3/4) ln(x^2 + 1) + (1/2) atan x + (3/2) ln (x - 1)

Final:
∫ (2x + 1) / ( (x^2 + 1) (x - 1) ) dx = (-3/4) ln(x^2 + 1) + (1/2) atan x + (3/2) ln (x - 1) + C

That concludes our integration techniques section. Next time we move to infinite limits.

Until next time, take care,

Eddie

This blog is property of Edward Shore. © 2012

Calculus Revisited #13: Integrals Involving Trigonometric Identities

We are at Part 13 of a 21 part series. This time we will work with integrals involving trigonometric identities.

Problems
All problems today will be indefinite integrals. Problems with definite integrals will be handled similarly.

Identities Involving
sin^2 x + cos^2 x = 1
cos (2x) = 1 - 2 sin^2 x
sin (2x) = 2 sin x cos x
sin^2 x = (1 - cos 2x)/2
cos^2 x = (1 + cos 2x)/2


1. ∫ cos^2 x dx

∫ cos^2 x dx
= ∫ 1 - sin^2 x dx
= ∫ 1 dx - ∫ sin^2 x dx
= x - ∫ 1/2 - 1/2 * cos 2x dx
= x - ( ∫ 1/2 dx - ∫ 1/2 * cos 2x dx )
= x - 1/2* x + 1/2 * ∫ cos 2x dx
= 1/2 * x + 1/2 * sin 2x * 1/2
= 1/2 * x + 1/4 * sin 2x

Final: ∫ cos^2 x dx = 1/2 * x + 1/4 * sin 2x + C

Note:
∫ cos (ax) dx = sin (ax)/a
∫ sin (ax) dx = -cos (ax)/a

2. ∫ sin^2 x cos^2 x dx

∫ sin^2 x cos^2 x dx
= ∫ (1 - cos 2x)/2 * (1 + cos 2x)/2 dx
= ∫ 1/4 * (1^2 - cos^2 (2x)) dx
= 1/4 * ( ∫ 1 - cos^2 (2x) dx )
= 1/4 * ( ∫ 1 - (1 + cos 4x)/2) dx )
= 1/4 * ( ∫ 1/2 - 1/2 * cos 4x dx)
= 1/4 * (1/2 * x - 1/8 * sin 4x)

Final: ∫ sin^2 x cos^2 x dx = 1/4 * (1/2 *x - 1/8 * sin 4x) + C

Trigonometric Substitutions
If the integrand has...

√(a^2 - x^2): use the substitution x = a sin u, dx = a cos u du

√(a^2 + x^2): use the substitution x = a tan u, dx = a sec^2 u du

√(x^2 - a^2): use the substitution x = a sec u, dx = a tan u sec u du

These can get messy!

Identity: 1 + tan^2 x = sec^2 x

3. ∫ 1 / √(4 - x^2) dx

a^2 = 4
a = 2
x = 2 sin u
dx = 2 cos u du

∫ 1/√(4 - x^2) dx
= ∫ (2 cos u) / √(4 - 4 sin^2 u) du
= ∫ (2 cos u) / (√4 * √(1 - sin^2 u) ) du
= ∫ (2 cos u) / (2 * √( cos^2 u ) ) du
= ∫ (2 cos u) / (2 * cos u) du
= ∫ 1 du
= u

x = 2 sin u implies that u = asin(x/2)

= asin(x/2)

Final: ∫ 1 / √(4 - x^2) dx = asin(x/2) + C

4. ∫ 1/ √(x^2 + 64) dx

a^2 = 64
a = 8
x = 8 tan u
dx = 8 sec^2 u du
u = atan(x/8)

∫ 1/ √(x^2 + 64) dx
= ∫ (8 sec^2 u)/(√(64 * tan^2 u + 64) du
= ∫ (8 sec^2 u)/(8 * √(sec^2 u)) du
= ∫ (8 sec^2 u)/(8 * sec u) du
= ∫ sec u du

∫ sec x dx = ln |sec x + tan x|

= ln |sec u + tan u|
= ln | sec(atan(x/8)) + tan(atan(x/8)) |

Refer to the right triangle below:

u = atan (x/8)

tan u = x/8

cos u = 8/√(x^2 + 64)

sec u = √(x^2 + 64)/8


Then:

ln | sec(atan(x/8)) + tan(atan(x/8)) |
= ln | √(x^2 + 64)/8 + x/8 |
= ln | √(x^2 + 64) + x | - ln 8

Final:
∫ 1 / √(x^2 + 64) dx
= ln | √(x^2 + 64) + x | - ln 8 + C

Note that ln 8 + C is a constant. Then.

∫ 1 / √(x^2 + 64) dx
= ln | √(x^2 + 64) + x | + C

5. ∫ 1 / (x^2 - 25)^(3/2) dx

a^2 = 25
a = 5
x = 5 sec u
u = asec(x/5) = acos(5/x)
dx = 5 sec u tan u du

1 + tan^2 x = sec^ x

Then:
∫ 1 / (x^2 - 25)^(3/2) dx
= ∫ (5 sec u tan u) / ( 25 sec^2 u - 25 )^(3/2) du
= ∫ (5 sec u tan u) / ( 25 tan^2 u )^(3/2) du
= ∫ (5 sec u tan u) / ( 25^(3/2) * tan^3 u ) du
= 5 / 25^(3/2) * ∫ (sec u tan u) / tan^3 u du
= 1/25 * ∫ cos u/sin^2 u du
= 1/25 * ∫ cot u csc u du

∫ cot x csc x dx = -csc x

5 / 25^(3/2) = 5 / ((5^2)^(3/2)) = 5 / (5^3) = 1 / 5^2 = 1/25

= 1/25 * -csc u
= -1/25 * csc(acos(5/x))

Refer to the triangle below:

x = 5 sec u

x/5 = sec u

5/x = cos u

√(x^2 - 25)/x = sin u

x/√(x^2 - 25) = csc u


Then:

-1/25 * csc(acos(5/x))
= -1/25 * x/√(x^2 - 25)

Final:
∫ 1 / (x^2 - 25)^(3/2) dx = -1/25 * x / √(x^2 - 25) + C

That concludes Part 13 of our series. Next time we will work with decomposition of fractions.

Have a good day,

Eddie


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