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 4: ∫x^2 / √(1 - x^2) dx

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Welcome to March Calculus Madness!

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∫x^2 / √(1 - x^2)  dx

Substitute:  z = arcsin x

sin z = x

sin^2 z = x^2

and dz = 1/√(1 - x^2) dx

Then:

∫x^2 / √(1 - x^2)  dx

= ∫ sin^2 z dz

= ∫  1/2 - 1/2 ∙ cos(2∙z) dz    

(by trigonometric identity of sin^2 z = 1/2 - 1/2 ∙ cos(2∙z))

= z/2 - 1/4 ∙ sin(2∙z) + C

= z/2 - 1/2 ∙ sin z ∙ cos z  + C

(by trigonometric identity of sin(2∙z) = 2 ∙ cos z ∙ sin z)

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

= 1/2 ∙ arcsin x - 1/2 ∙ x ∙ √(1 - x^2) + C

(sin(arcsin x) = x, cos(arcsin x) = √(1 - x^2)

Summary:

∫x^2 / √(1 - x^2)  dx  = 1/2 ∙ arcsin x - 1/2 ∙ x ∙ √(1 - 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. 

Back to School: Survival Sheet for Algebra and Pre-Calculus (Trigonometry)

Back to School:  Survival Sheet for Algebra and Pre-Calculus (Trigonometry)

These are common topics found in algebra and pre-calculus classes.  You can download a pdf version of each.  I hope you find this helpful not only in classes but beyond.

Algebra:  https://drive.google.com/open?id=0B7R8x9Yi26yGQ1g4NzZOcm43Zm8

Pre-Calculus/Trigonometry:  https://drive.google.com/open?id=0B7R8x9Yi26yGem9sQ1JfSGplOWc

Algebra

Pre-Calculus/Trigonometry

This blog is property of Edward Shore, 2016.  

HP Prime: Trigonometric CAS Rewrite Commands

HP Prime:  Trigonometric CAS Rewrite Commands

All of the following examples have CAS Simplification set to Minimum.

Rewrites of Arcsine and Sine

asin2acos	Toolbox → 4. Rewrite → 5. Sine → 1. asin x → acos x	asin2acos(2*asin(x)) returns 2*(-acos(x) + π/2)
asin2atan	Toolbox → 4. Rewrite → 5. Sine → 2. asin x → atan x	asin2atan(2*asin(x)) returns 2*(atan(x/√(-x^2+1))
sin2costan	Toolbox → 4. Rewrite → 5. Sine → 3. sin x → cos x tan x	sin2costan(2*sin(x)^2) returns 2*(cos(x)*tan(x))^2

Rewrites of Arcosine and Cosine

acos2asin	Toolbox → 4. Rewrite → 6. Cosine → 1. acos x → asin x	acos2asin(acos(x*π/2)) returns -asin(1/2*π*x)+π/2
acos2atan	Toolbox → 4. Rewrite → 6. Cosine → 2. acos x → atan x	acos2atan(acos(x*π/2)) returns -atan( (1/2*π*x)/ (√(1/4) – √(-(π*x)^2+4) )+π/2
cos2sintan	Toolbox → 4. Rewrite → 6. Cosine → 3. cos x → sin x/tan x	cos2sintan(2*cos(x)^2) Returns 2*(sin(x)/tan(x))^2

Rewrites of Arctangent and Tangent

atan2asin	Toolbox → 4. Rewrite → 7. Tangent → 1. atan x → asin x	atan2asin(atan(x/3)) returns asin( (x/3) / (√(1/9)*√(x^2+9)) simplify gets asin( (x*√(x^2+9)) / (x^2+9) )
atan2acos	Toolbox → 4. Rewrite → 7. Tangent → 2. atan x → acos x	atan2acos(atan(x/3)) returns -acos( (x/3) / (√(1/9)*√(x^2+9))+π/2 simplify gets ( (π-2*acos( (x*√(x^2+9))/(x^2+9) )/2
tan2sincos	Toolbox → 4. Rewrite → 7. Tangent → 3. tan x → sin x/cos x	tan2sincos(tan(2*x)) returns sin(2*x)/cos(2*x)
halftan (sin x, cos x, tan x) → tan(x/2)	Toolbox → 4. Rewrite → 7. Tangent → 4. halftan	halftan(cos(2*x)) returns (-tan(x)^2+1)/ (tan(x)^2+1)

Other Commands

tlin (trigonometric form of powers) → (trigonometric form of linearized terms)	Toolbox → 4. Rewrite → 8. Trig → 5. tlin	tlin(cos(2*x)^2) returns 1/2*cos(4*x)+1/2
trig2exp (sin x, cos x, tan x) → (exponential form with complex exponentials)	Toolbox → 4. Rewrite → 8. Trig → 8. trig2exp	trig2exp(sin(x/3)) returns 1/(2*i) * (e^(i*x/3) – 1/e^(i*x/3))
atrig2ln (asin x, acos x, atan x) → (logarithm form with complex exponentials)	Toolbox → 4. Rewrite → 8. Trig → 4. atrig2ln	atrig2ln(asin(x/3)) returns i*ln(√(1/9)*√(x^2-9)+x/3)+π/2

That is a list of basic CAS commands for trigonometric rewrites. 

Eddie

This blog is property of Edward Shore, 2016.

HP Prime: Algebraic CAS Commands

HP Prime:  Algebraic CAS Commands

On today’s entry I am going discuss several CAS (Computer Algebraic System) commands: factor, expand, collect, and subst (substitute).   These four commands are common to calculators, mathematical software, and mathematical apps that have CAS commands.

If you have not heard about CAS before, CAS allows the user to operate on expressions with indefinite (undefined) variables.  CAS commands commonly manipulate algebraic expressions, calculate derivatives of functions, and calculate indefinite integrals.    My first experience with CAS commands was with the (original) TI-89 almost a decade and half ago.  (Wow, I’m getting old).  Let’s get started. 

On the HP Prime, you can enter CAS mode by pressing the [CAS] button. In CAS mode, the variables are primarily lower case.  Single letter uppercase variables, along with θ, are always numeric. 

Note:  Radian mode is used.

Factor

Two type of factoring functions:  factor and ifactors.

factor:  [Toolbox], (CAS), 1. Algebra, 4. Factor

The command factor works with polynomials and rational expressions. 

Examples: 

factor(x^2 + 6*x + 5) returns (x + 1)*(x + 5)

factor(x^2 + x – 4) returns (x + (1 - √17) / 2) * (x + (1 + √17) / 2)

factor(x^2 + 4*x + 4) returns (x + 2)^2

TIP:  If you are not sure whether implied multiplication will be accepted, insert the multiply operators (*) in the expression.  It is always a safe bet. 

ifactors:  [Toolbox], (CAS), 5.  Integer, 2. Factors

Unlike factor, ifactors works on factoring integers into their prime factorizations.  If the integer is prime, the integer itself is returned. 

Examples:

ifactors(55) returns 5 * 11

ifactors(2162) returns 2 * 23 * 47

ifactors(1367) returns 1367   (1367 is prime)

Expand

There are three expand commands we’ll cover:  expand, powexpand, and texpand.  Which command that you use determines what time of expression to be expanded. 

expand:  [Toolbox], (CAS), 1.  Algebra, 4.  Factor

The command expand works on polynomials and rational functions.  It can be thought of that expand is the opposite of factor. 

Examples:

expand( (x - 1)^3 ) returns x^3 – 3*x^2 + 3*x - 1

expand( (x + 1)^2 * (x^2 – 3) ) returns x^4 + 2*x^3 – 2*x^2 – 6*x – 3

powexpand:  [Toolbox], (CAS), 4.  Rewrite, 2.  powexpand

The command powexpand involves expressions with exponents. 

powexpand( x^(2 + a) ) returns x^2 * x^a

Sometimes you have to combine one or more CAS commands to get what you want. 

powexpand( x^((a + 1) * (a + 5)) ) returns x^((a + 1) * (a + 5)).   

This is not a result I would want.  Let’s tackle this another way: 

expand( (a + 1)*(a + 5) ) returns a^2 + 6*a + 5, then

powexpand( x^(a^2 + 6*a + 5) ) returns x^a^2 * (x^a)^6 * x^5

texpand:  [Toolbox], (CAS), 4. Rewrite, 3. Texapnd

The command texpand is part of the expand family, this time texpand works with transcendental functions (sin, cos, tan, ln, exp). 

texpand( sin(2*x) ) returns 2 * cos(x) * sin(x)

texpand( cos (2*x + 1) ) returns (2 * cos(x)^2 – 1) * cos(1) – 2 * cos(x) * sin(x) * sin(1)

texpand( e^(3*x) + cos(3*x) ) returns e^x^3 + 4 * cos(x)^3 – 3 * cos(x)

Collect

collect:  [Toolbox], (CAS), 1.  Algebra, 2.  Collect

This commands collects like terms in a polynomial, and when applicable factorizes the expressions.  The variable to collect around may be specified.

Examples:

collect( (x + 1)^2 + x^2 + 2*x ) returns 2*x^2 + 4*x + 1

collect( (x + a + 2*(x + a) ) returns 3 * (a + x)

collect( (x + a + 2*(x + a), a ) returns 3*a + 3*x

collect( (x + a + 2*(x + a), x ) returns 3*a + 3*x

collect( (x^2 + a^2 + 2*(x + a)^2 ) returns 4*a*x + 3*a^2 + 3*x^2

collect( (x^2 + a^2 + 2*(x + a)^2, a ) returns a * (3*a + 4*x) + 3*x^2

collect( (x^2 + a^2 + 2*(x + a)^2, x ) returns x * (4*a + 3*x) + 3*a^2

Substitute

This does exactly what it is says:  substitute.  

Syntax:  subst(expression, var = value/expr)

subst:  [Toolbox], (CAS), 1. Algebra, 6. Substitute

Examples:

subst( x^4 + 2*x^2 – x, x = √a) returns a^2 + 2*a - √a

Here are two examples where an additional CAS command is needed to simplify:

subst( cos(2*x), x = asin(a) ) returns cos( 2* asin(a) )

texpand(cos( 2* asin(a) ) returns 2*(1 – a^2) – 1

subst( 2*x^2 – x, x = a + 3 ) returns 2*(a + 3)^2 – a – 3

expand(2*(a + 3)^2 – a – 3 ) returns 2*a^2 + 11*a + 15

There are four common CAS commands, I hope you find this helpful.

Eddie

This blog is property of Edward Shore, 2016.

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


This blog is property of Edward Shore. © 2012