Fun With The HP 27S

Fun With The HP 27S

Notes:

*  The HP 27S is set to FIX 4

*  Since none of the formulas have trigonometric functions, they can also be programmed on the HP 17B and HP 17BII+

Partial Fraction Decomposition

(A ∙ x + B)÷((x + C) ∙ (x + D)) = R÷(x + C) + S÷(x + D)

Inputs:  A, B, C, D

Outputs:  R, S

Formula:

PARTFRAC2:(A+B+C+D)×0=IF(S(R):(B-C×A)÷(D-C)-R:(B-D×A)÷(C-D)-S)

Example:  (4 ∙ x + 3) ÷ ((x - 5) ∙ (x + 1))

A = 4, B = 3, C = -5, D = 1

Results:  R = 3.8333, S = 0.1667

Use that the 0×(var1+var2+...) to set an order of variables in the solver.  

2 x 2 Simultaneous Equations

A ∙ x + B ∙ y = C

D ∙ x + E ∙ y = F

The solutions are:

x = (C ∙ E - B ∙ F) ÷ (A ∙ E - B ∙ D)

y = (A ∙ F - C ∙ D) ÷ (A ∙ E - B ∙ D)

Formula:

SIM2X2:0×(A+B+C+D+E+F+L(M:A×E-B×D))=IF(S(X):(C×E-B×F)÷G(M)-X:(A×F-C×D)÷G(M)-Y)

Example:  2x + 3y =5, -3x + 8y = -8

A = 2, B = 3, C = 5, D = -3, E = 5, F = -8

Results:  X = 2.5789, Y = -0.0526

Floor Function

floor(x):  the greatest integer less than or equal to x

floor(x):

If frac(x) = 0 Then return x

 Else If x≥0, then return intg(x) else return intg(x)-1

Formula:

FLOOR=IF(FP(X)=0:X:IF(X>=0:IP(X):IP(X)-1))

Examples:

X = 2.38, FLOOR = 2.0000

X = -9.21, FLOOR = -10.0000

Ceiling Function

ceil(x):  the least integer greater than or equal to x

ceil(x):

If frac(x) = 0 Then return x

  Else If x≥0, then intg(x)+1 else return intg(x)

Formula:

CEIL=IF(FP(X)=0:X:IF(X>=0:IP(X)+1:IP(X)))

Examples:

X = 2.38, CEIL = 3.0000

X = -9.21, CEIL = -9.0000

Rydberg Formula

The Rydberg formula measures the light's wavelength when an electron moves between energy quantum numbers (from higher to lower levels).  The Rydberg formula is to be used for simple atoms only, and is most used for hydrogen atoms.

1/λ = R ∙ Z^2 ∙ (1/n1^2 - 1/n2^2)

R = Rydberg's Constant ≈ 1.097373157 × 10^7 m^(-1)

Z = atomic number, 1 for hydrogen

n1, n2:  energy levels

Formula:

RYDBERG:INV(L)=1.09737316E7×SQ(Z)×(INV(SQ(N1))-INV(SQ(N2)))

N2 > N1

Example:

The energy of an hydrogen election from level 4 to level 2.  Z = 1

Z =1,  N1 = 2, N2 = 4

Result:  L = 4.8601E-7

Source:

Helmenstine, Todd. "What Is the Rydberg Formula and How Does It Work?" ThoughtCo, Aug. 28, 2020, https://www.thoughtco.com/what-is-the-rydberg-formula-604285   Retrieved November 4, 2021

Moment of Inertia - Circular Ring

D$OUT:  outside diameter

D$IN:  insider diameter.  For a circle, set D$IN = 0

I: moment of inertia

Formula:

I=PI×(D$OUT^4-D$IN^4)÷64

Examples:

Circular Ring:

D$OUT = 6.2, D$IN = 1.9;  I = 71.8935

Circle:

D$OUT = 6.2, D$IN = 0; i = 72.5332

Source:

"Properties Of Annual Sections" HP-19C/HP-29C Solutions: Civil Engineering  Hewlett-Packard.  1977

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.

12 Days of Christmas Integrals: ∫ x^n ∙ e^x dx; n is a positive integer

12 Days of Christmas Integrals:  ∫ x^n ∙ e^x dx;  n is a positive integer 

On the Twelfth day of Christmas Integrals, the integral featured today is...

∫ x^n ∙ e^x dx;  n is a positive integer 

n = 1, 2, 3, 4, ...

Each term is found by repeating the Integration by Parts method.  Today, we will find a (somewhat) closed formula that applies to any positive integer n.  

∫ x^n ∙ e^x dx

u = x^n

du = n ∙ x^(n-1) dx

dv = e^x dx

v = e^x

= x^n ∙ e^x - ∫ n ∙ x^(n-1) ∙ e^x dx

u = n ∙ x^(n-1)

du = n ∙ (n-1) ∙ x^(n-2) dx

dv = e^x dx

v = e^x

= x^n ∙ e^x - n ∙ x^(n-1) ∙ e^x + ∫ n ∙ (n-1) ∙ x^(n-2) ∙ e^x dx

u = n ∙ x^(n-1)

du = n ∙ (n-1) ∙ x^(n-2) dx

dv = e^x dx

v = e^x

and so on until...

u = n ∙ (n - 1) ∙ ... ∙ 2 ∙ x

du = n! dx

dv = e^x dx

v = e^x

with the closed form:

= e^x ∙ [x^n - n ∙ x^(n-1) + n ∙ (n-1) ∙ x^(n-2) - n ∙ (n-1) ∙ (n-2) ∙ x^(n-3) + ...  

   + (-1)^(n mod 2) ∙ n!]

Examples: 

n = 2

∫ x^2 ∙ e^x dx = e^x ∙ (x^2 - 2 ∙ x + 2)

n = 3

∫ x^3 ∙ e^x dx = e^x ∙ (x^3 - 3 ∙ x^2 + 6 ∙ x - 6)

n = 4

∫ x^4 ∙ e^x dx = e^x ∙ (x^4 - 4 ∙ x^3 + 12 ∙ x^2 - 24 ∙ x + 24)

What do you think of the integral series?  Is this worth a sequel? 

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. 

12 Days of Christmas Integrals: ∫ (a ∙ b) ÷ (a^2 ∙ x^2 + b^2) dx; a,b are constants

 12 Days of Christmas Integrals:  ∫ (a ∙ b) ÷ (a^2 ∙ x^2 + b^2) dx; a, b are constants

On the Eleventh day of Christmas Integrals, the integral featured today is...

∫ (a ∙ b) ÷ (a^2 ∙ x^2 + b^2) dx; a, b are constants

A rather simple integral:

∫ (a ∙ b) ÷ (a^2 ∙ x^2 + b^2) dx

= a ∙ b ∙ ∫ 1÷ (a^2 ∙ x^2 + b^2) dx

= a ∙ b ∙ ∫ 1÷ (b^2 ∙ (a^2/b^2 ∙ x^2 + 1)) dx

= a/b ∙ ∫ 1÷ (a^2/b^2 ∙ x^2 + 1) dx

= ∫ (a/b) ÷ ((a/b)^2 ∙ x^2 + 1) dx

= arctan(a/b ∙ x) + C

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. 

12 Days of Christmas Integrals: ∫ x^2 ∙ exp(x)^2 dx

 12 Days of Christmas Integrals:  ∫ x^2 ∙ exp(x)^2 dx   

On the Tenth day of Christmas Integrals, the integral featured today is...

∫ x^2 ∙ exp(x)^2 dx   

exp(x) = e^x

Start with the substitution:

u = exp(x)

du = exp(x) dx

1 ÷ exp(x) du = dx

1 ÷ u du = dx 

u^2 = exp(x)^2

ln(u) = x

ln(u)^2 = x^2

Then:

∫ x^2 ∙ exp(x)^2 dx   

= ∫ ln(u)^2 ∙ u^2 ∙ 1/u du

= ∫ u ∙ ln(u)^2 du

Note:  We did this integral on the Seventh Day of 12 Days of Christmas integrals (December 31, 2021).  Check out that post for details!

= u^2/2 ∙ ln(u)^2 - u^2/2 ∙ ln(u) + u^4/4 + C

Recall u = exp(x):

= 1/2 ∙ (exp(x))^2  ∙ x^2 - 1/2 ∙ (exp(x))^2 ∙ x + 1/4 ∙ (exp(x))^2 + C 

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. 

12 Days of Christmas Integrals: ∫ 1 ÷ (√(x + a) - √(x)) dx, a is a constant

 12 Days of Christmas Integrals:  ∫ 1 ÷ (√(x + a) - √(x)) dx, a is a constant

On the Ninth day of Christmas Integrals, the integral featured today is...

∫ 1 ÷ (√(x + a) - √(x)) dx, a is a constant  

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

= ∫ 1 ÷ (√(x + a) - √(x)) ∙ (√(x+a) + √(x)) ÷ (√(x+a) + √(x)) dx

= ∫ (√(x+a) + √(x)) ÷ (x + a - x) dx

= ∫ (√(x+a) + √(x)) ÷ (a) dx

= ∫ √(x + a) ÷ a + √(x) ÷ a dx

= 2/3 ∙ (x + a)^(3/2) + 2/3 ∙ x^(3/2) ÷ a

= 2 ÷ (3 ∙ a) ∙ ((x + a)^(3/2) + x^(3/2))

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. 

12 Days of Christmas Integrals: ∫ (x^2 - 4 ∙ x + 3) ÷ (x^2 - 6 ∙ x + 8) dx

 12 Days of Christmas Integrals:  ∫ (x^2 - 4 ∙ x + 3) ÷ (x^2 - 6 ∙ x + 8) dx

HAPPY NEW YEAR!!!!   HERE'S TO A WONDERFUL 2022!  May we make our future and present bright!  (We all could use a bright year, given how the last few years went)

On the Eighth day of Christmas Integrals, the integral featured today is...

∫ (x^2 - 4 ∙ x + 3) ÷ (x^2 - 6 ∙ x + 8) dx

By division:

= ∫ 1 + (2 ∙ x - 5) ÷ (x^2 - 6 ∙ x + 8) dx

Find the partial fraction of (2 ∙ x - 5) ÷ (x^2 - 6 ∙ x + 8) 

(2 ∙ x - 5) ÷ (x^2 - 6 ∙ x + 8) = A ÷ (x - 2) + B ÷ (x - 4)

2 ∙ x - 5 = A ∙ (x - 4) + B ∙ (x - 2)

2 ∙ x - 5 = A ∙ x - 4 ∙ A + B ∙ x - 2 ∙ B

which implies that:

-5 = -4 ∙ A - 2 ∙ B

2 = A + B

Solving for A and B:

A  = 1/2 and B = 3/2 

(2 ∙ x - 5) ÷ (x^2 - 6 ∙ x + 8) = (1/2) ÷ (x - 2) + (3/2) ÷ (x - 4)

= ∫ 1 + (1/2) ÷ (x - 2) + (3/2) ÷ (x - 4) dx

= x + 1/2 ∙ ln|x - 2| + 3/2 ∙ ln|x - 4| + C

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. 

12 Days of Christmas Integrals: ∫ x ∙ (ln(x))^2 dx

12 Days of Christmas Integrals:  ∫ x ∙ (ln(x))^2 dx

NEW YEARS EVE!!!!

On the Seventh day of Christmas Integrals, the integral featured today is...

∫ x ∙ (ln(x))^2 dx

Sounds like a job for integration by parts!

∫ x ∙ (ln(x))^2 dx

u = (ln(x))^2 

du = 2 ∙ ln(x) ∙ 1/x dx

dv = x dx

v = x^2/2

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

= x^2/2 ∙ (ln(x))^2 - ∫ x ∙ ln(x) dx

u  = ln(x)

du = 1/x dx

dv = x dx

v = x^2/2

= x^2/2 ∙ (ln(x))^2 - x^2/2 ∙ ln(x) + ∫ x/2 dx

= x^2/2 ∙ (ln(x))^2 - x^2/2 ∙ ln(x) + x^2/4 + C

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. 

12 Days of Christmas Integrals: ∫ cos (√x) dx

12 Days of Christmas Integrals:  ∫ cos (√x) dx

On the Sixth day of Christmas Integrals, the integral featured today is...

∫ cos (√x) dx

Handling this integral will require two integral methods.  First substitution:

Let u = √x = x^(1/2)

Then:

du = 1/2 ∙ x^(-1/2) dx

2 ∙ x^(1/2) du = dx

2 ∙ u du = dx

∫ cos(√x) dx

= ∫ cos(x^(1/2)) dx

= ∫ 2 ∙ u ∙ cos(u) du

At this point, we now apply Integration by Parts:

w = 2 ∙ u

dw = 2 du

dv = cos(u) du

v = sin(u)

= 2 ∙ u ∙ sin(u) - ∫ 2 ∙ sin(u) du

= 2 ∙ u ∙ sin(u) +  2 ∙ cos(u) + C

Recall u = x^(1/2):

= 2 ∙ x^(1/2) ∙ sin(x^(1/2)) + 2 ∙ cos(x^(1/2)) + C

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. 

12 Days of Christmas Integrals: ∫ 2 ∙ x ∙ (x^2 + 1)^2 dx

12 Days of Christmas Integrals:  ∫ 2 ∙ x ∙ (x^2 + 1)^2 dx

On the Fifth day of Christmas Integrals, the integral featured today is...

∫ 2 ∙ x ∙ (x^2 + 1)^2 dx

I am going to use two approaches and show why both approaches are valid.  Remember we are working indefinite integrals.  

Substitution Method

∫ 2 ∙ x ∙ (x^2 + 1)^2 dx

u = x^2 + 1 

du = 2 ∙ x dx

u^2 = (x^2 + 1)^2

∫ u^2 du

= u^3 ÷ 3 + C

= (x^2 + 1)^3 ÷ 3 + C

= (x^6 + 3 ∙ x^4 + 3 ∙ x^2 + 1) ÷ 3 + C

= (x^6 + 3 ∙ x^4 + 3 ∙ x^2) ÷ 3 + 1 ÷ 3 + C

= x^6 ÷ 3 +  x^4 +  x^2 + 1 ÷ 3 + C

Algebraic Method

∫ 2 ∙ x ∙ (x^2 + 1)^2 dx

= ∫ 2 ∙ x ∙ (x^4 + 2 ∙ x^2 + 1) dx

= ∫ 2 ∙ x^5 + 4 ∙ x^3 + 2 ∙ x dx

= 2/6 ∙ x^6 + 4/4 ∙ x^4 + 2/2 ∙ x^2 + C

= x^6 ÷ 3 + x^4 + x^2 + C

Why are the two methods acceptable?

In indefinite integration, a "+ C" is added to show that the solutions to indefinite integrals are a class of functions.   C is an arbitrary numerical constant.

d/dx[  f(x) + C ] =   d/dx [ f(x) ] + 0  = d/dx [ f(x) ]

Since C is a numerical constant,  C + 1/3 is also a numerical constant.  

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. 

12 Days of Christmas Integrals: ∫ x^2 ÷ (1 + x^2) dx

 12 Days of Christmas Integrals:  ∫ x^2 ÷ (1 + x^2) dx

On the Fourth day of Christmas Integrals, the integral featured today is...

∫ x^2 ÷ (1 + x^2) dx

Time for a little algebraic manipulation:  add and subtract 1 ÷ (1 + x^2) 

= ∫ x^2 ÷ (1 + x^2) + 1 ÷ (1 + x^2) - 1 ÷ (1 + x^2)   dx

= ∫ (x^2 + 1) ÷ (1 + x^2) - 1 ÷ (1 + x^2)   dx

= ∫ (1 + x^2) ÷ (1 + x^2) - 1 ÷ (1 + x^2)   dx

= ∫ 1 - 1 ÷ (1 + x^2)   dx

= x - arctan(x) + C

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. 

12 Days of Christmas Integrals: ∫ (sin(x))^2 dx

 12 Days of Christmas Integrals:  ∫ (sin(x))^2 dx

On the Third day of Christmas Integrals, the integral featured today is...

∫ (sin(x))^2 dx

This calls for the trig identity:

(sin(x))^2 = 1/2 - 1/2 ∙ cos(2 ∙ x)

With any operations in calculus involving trigonometric functions, the angle units are assumed to be in radians.

∫ (sin(x))^2 dx 

= ∫ 1/2 - 1/2 ∙ cos(2 ∙ x) dx

= 1/2 ∙ x - 1/4 ∙ sin(2 ∙ x) + C

= 1/2 ∙ x - 1/2 ∙ sin(x) ∙ cos(x) + C

Simple as that.   We will get meatier integrals as the days wear on.  

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. 

12 Days of Christmas Integrals: ∫ (x ∙ arctan(x)) ÷ (1 + x^4) dx

 12 Days of Christmas Integrals:  ∫ (x ∙ arctan(x)) ÷ (1 + x^4) dx

On the Second day of Christmas Integrals, the integral featured today is...

∫ (x ∙ arctan(x)) ÷ (1 + x^4) dx

This integral can be approached by using the substitution method:  

Let u = arctan(x^2).  Then:

du = 1 ÷ (1 + x^4) ∙  d/dx(x^2)

du = 1 ÷ (1 + x^4) ∙  (2  ∙ x) dx

1/2 du = x ÷ (1 + x^4) dx

Then:

∫ (x ∙ arctan(x)) ÷ (1 + x^4) dx

= ∫ u ∙ 1/2 du

= 1/2 ∙ ∫ u du

 = u^4/4 + C

Substitute back:

= (arctan(x^2))^2 / 4 + C

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. 

12 Days of Christmas Integrals: ∫ π ∙ cos(x) ∙ sin(x) dx

 12 Days of Christmas Integrals:  ∫ π ∙ cos(x) ∙ sin(x) dx

In the spirit of the Christmas Holiday, I am presenting you with the 12 Days of Christmas Integrals!

On the First day of Christmas Integrals, the integral featured today is...

∫ π ∙ cos(x) ∙ sin(x) dx

Let's start off with using the trigonometric identity

sin(2 ∙ x) = 2 ∙ cos(x) ∙ sin(x)

1/2 ∙ sin(2 ∙x) = cos(x) ∙ sin(x)

Then:

∫ π ∙ cos(x) ∙ sin(x) dx

= π ∙ ∫ 1/2 ∙ sin(2 ∙ x) dx

= π/2 ∙ ∫ sin(2 ∙ x) dx

Multiply by both 1/2 and 2:

= π/4 ∙ ∫ 2 ∙ sin(2 ∙ x) dx

= -π/4 ∙ cos(2 ∙ x) + C 

[ ∫ sin x dx = -cos x + C,  angles are in radians ]

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.