Sec 5.8 Page 386 # 53

y = (1/2)*[(1/2)Ln ((x+1)/(x-1)) + arctan x]

Things needed to do the problem.

(d/dx)[arctan u] = [u' / (1 + u^{^2})]

Let u = x,  then u ' = 1

therefore: 

(d/dx)[arctan u] = [1 / (1 + x^{^2})]

More things needed to do the problem.

(d/dx)[Ln x] = (1/x)

Simplify the problem, use rules about Natural Logs

Also factor out the constant in front of the Ln..

y = (1/2)*(1/2)*[(Ln(x+1) - Ln(x-1)] + [(1/2) arctan x]

Now diff of eqn:

y ' = (1/4)*[1/(x+1) - 1/(x-1)] +  [(1/2) * (1 / (1 + x^{^2}))]

Get a common denominator so terms can be combined  

Common denominator 1st part

[(1/(x+1)) * (x-1)] = [(x-1) / (x^{^2}-1)]

Common denominator 2nd part

[(1/(x-1)) * (x+1)] = [(x+1) / (x^{^2}-1)]

Set back into equation.  

y ' = (1/4)*[((x-1)/(x^{^2}-1)) - ((x+1)/(x^{^2}-1))] +  [(1/2)*(1 /  (x^{^2} + 1))] 

Then combine and simplify

(1/4)*[-2/(x^{^2}-1))] +  [(1/2)*(1 / (x^{^2} + 1))]

Simplify the problem more.

(-1/2)*[(1/(x^{^2}+1))] +  [(1/2)*(1 / (x^{^2} + 1))]

Factor out (-1/2) to simplify

(-1/2)*[(1/(x^{^2}-1)) +  (1 / (x^{^2} + 1))]

Get a common denominator so terms can be combined 

 (x^{^4} - 1)

Get a common denominator of 1st part

(1/(x^{^2}-1)*[ (x^{^2}+1)/(x^{^2}+1)] 

= (x^{^2}+1)/ (x^{^4} - 1)

Get a common denominator of 2nd part

(1 / (x^{^2} + 1))*[ (x^{^2}-1)/(x^{^2}-1)]

= (x^{^2}-1)/(x^{^4} - 1)

(-1/2)*[((x^{^2}+1)/ (x^{^4} - 1)) +  ((x^{^2}-1)/(x^{^4} - 1))]

(-1/2)*[((x^{^2}+1)*(x^{^2}-1))/ (x^{^4} - 1)]

(-1/2)*[(-2)/ (x^{^4} - 1)] 

Final solution:

y ' (x) = [1/(x^{^4} - 1)] 

Sec 5.8 Page 386 # 57

y = 8 arcsin( x / 4) - [(x*(sqrt (16 - x^{^2})))/2]

Simplify the problem, change sqrt into exp of (1/2)

this makes it easier to use the chain rule.

Things needed to do the problem.

(d/dx)[arcsin u] = [u' / (sqrt (1-u^{^2}))]

Let u = ( x / 4),  then u ' = (1/4)

therefore: 

(d/dx)[arcsin u] = [(1/4) / (sqrt (1 - ( x / 4)^{^2}))]

y ' (x) = [8 * (d/dx)(arcsin( x / 4))] - [((d/dx)x*(16 - x^{^2})^{^(1/2)}))/2  ]+[(x*(d/dx)(16 - x^{^2})^{^(1/2)}))/2  ]

y ' (x) = [8 *[(1/4) / (sqrt (1 - ( x / 4)^{^2}))]] - [ (1)*((16 - x^{^2})^{^(1/2)}))/2) + (x/2)*(1/2)*(-2x)*((16 - x^{^2})^{^(-1/2)})]

Note, the derivative of 8 is zero therefore that part was not shown here since it would have cancelled out.

Now you need to simplify the problem as much as possible for the next few steps.

y ' (x) = [2* (1/ (sqrt (1 - ( x / 4)^{^2}))] - [(sqrt (16 - x^{^2})))/2] - [(x/4)*(-2x)*((16 - x^{^2})^{^(-1/2)}) ] 

y ' (x) = [2* (1/ (sqrt (1 - ( x ^{^2})/ 16)))] - [(sqrt (16 - x^{^2})))/2] + [(x^{^2}) /((2)*(sqrt (16 - x^{^2}))) ] 

y ' (x) = [2* (1/ (sqrt (16 - ( x ^{^2}))/ 16)))] - [(sqrt (16 - x^{^2})))/2] + [(x^{^2}) /((2)*(sqrt (16 - x^{^2}))) ] 

y ' (x) = [4* (1/ (sqrt (16 - ( x ^{^2})))] - [(sqrt (16 - x^{^2})))/2] + [(x^{^2}) /((2)*(sqrt (16 - x^{^2}))) ] 

Final solution:

y ' (x) = ( x ^{^2}) /(sqrt (16 - ( x ^{^2})) 

This spot is reserved for problem 73

Coming soon!!!

Sec 5.8 Page 387 # 76

Verify each differentiation formula.

All of these are shown on the Visual Basic site:

Visual Basic- This shows how they are derived.

This is the last of the examples from Sec 5.8

Sec 5.9 Page 393 # 25

Things needed to do the problem.

 ∫ (1 / ((x)^{^(1/2)}*(1-x)^{^(1/2)}))dx

You need to recognize the pattern form(s) of that equation.

If we let u = (x)^{^(1/2)} 

then du = 1/(2* (x)^{^(1/2)})dx 

or 

dx = (2* (x)^{^(1/2)})du

If x = u² , then u =  ((u²)^{^(1/2)}) 

this makes

dx = (2 * u)du

  The NEW form of the equation is like this:              

∫ ((2 * u)du / (u*(1-u²)^{^(1/2)})

Cancel u in denominator and numerator

and move constant (2) out front of integral

This simplifies to :

 2* ∫ (du  /(1-u²)^{^(1/2)})    

Now it  matches this form:

∫ (du / (a² - u² )^{^(1/2)})  = arcsin( u / a ) + C  

 2* ∫ (du /(1-u²)^{^(1/2)})  

  = 2*arcsin( u / a ) + C 

   = 2*arcsin( (x)^{^(1/2)} / 1 ) + C 

This can be simplified for final answer.

Final solution:

  = 2*arcsin (x)^{^(1/2)}  + C 

Sec 5.9 Page 393 # 29

Things needed to do the problem.

∫ ((x + 5)*dx / (9 - ( x -3)²)^{^(1/2)}))

You need to recognize the pattern form(s) of that equation.

This equation  matches this form:

∫ (du / (a² - u² )^{^(1/2)})  = arcsin( u / a ) + C   

If you let a = 3 and u = ( x -3)

Since u = ( x - 3),  then du = 1 

But here would the numerator would be

(x + 5)* dx  so something more must be done.

We need ( x - 3) du  * dx 

Since (x + 5 ) - 8 = (x -3) we can use that.

 or      (x -3) + 8 = (x + 5 )

∫ ((x -3)*dx / (9 - ( x -3)²)^{^(1/2)}))  +  ∫ (8dx / (9 - ( x -3)²)^{^(1/2)}))

∫ (8 dx/ (9 - ( x -3)²)^{^(1/2)}))  

You can find this integral (it fits the Arcsin Rule)

∫ (du / (a² - u² )^{^(1/2)})= arcsin( u / a ) + C 

 or 8 arcsin (( x -3)/3) + C

 ∫ ((x -3)dx  / (9 - ( x -3)²)^{^(1/2)}))

You can find this integral ( it fits the Power Rule)

 or   - (6x - x² )^{^(1/2)})

= - (6x - x² )^{^(1/2)}) + 8 arcsin (( x -3)/3) + C

8 arcsin (( x -3)/3) + C 

 can be simplified to

 8 arcsin (( x /3)+ 1) + C

Final solution:

=  - (6x - x² )^{^(1/2)}) + 8 arcsin (( x /3)+ 1) + C

Sec 5.9 Page 393 # 33

Things needed to do the problem.

∫ ((2x)dx / ( x ² + 6x + 13))

If  u = ( x ² + 6x + 13)

then du = (2x + 6)  

but we have (2x),

therefore we must add and 

subtract 6 in the numerator.

∫ ((2x + 6 - 6)dx / ( x ² + 6x + 13))

This is not in a form where we can find the integral.

∫ ((2x + 6)dx / ( x ² + 6x + 13)) + ∫ ((-6)* dx/ ( x ² + 6x + 13))  

OR

∫ ((2x + 6)dx / ( x ² + 6x + 13)) - 6 ∫ (dx/ ( x ² + 6x + 13))

This is getting closer but the second equations

- 6 ∫ ((1)* dx/ ( x ² + 6x + 13))  or

- 6 ∫ ((dx) / ( x ² + 6x + 13))

 needs  some more adjustment to matches the form:

∫ ((du) / (a²   + u²  ))  = (1/a)* arctan (u/a) + C

This one will require that you complete the square.

( x ² + 6x + 13) --> you need  6/2 = 3 then 3²  = 9.

You must add 9 and subtract 9.

 = ( x ² + 6x + 9) - 9 + 13

 = ( x ² + 3)²  - 9 + 13

= ( x ² + 3)²  + 4

= - 6 ∫ (dx / (( x ² + 3)²  + 4))  or

= - 6 ∫ (dx / ( 4 + ( x ² + 3)²  ))

In this case:  a = 4, u = ( x ² + 3)

∫ (((2x + 6)*dx) / ( x ² + 6x + 13) )) - 6∫ (dx / (4 + ( x  + 3)²  ))

Now we can get the integral for both parts of the whole equation.           

The 1st part of the equations

∫ (((2x + 6)*dx) / ( x ² + 6x + 13) ))   

= Ln| x ² + 6x + 13| 

The 2nd part of the equations

- 6∫ (((1)*dx) / (4 + ( x  + 3)²  ))  

= - 6arctan((x+3)/4) + C

Together the whole equation is

  = Ln| x ² + 6x + 13| - 6arctan((x+3)/4) + C    

Final solution:

After simplification:

    = Ln| x ² + 6x + 13| - 3arctan((x+3)/2) + C

This is the end of the problems for assignment in Section 5.9

If you want others added, e-mail the problems to me and I will post them.

Problems 15,17,19,21,23

Problems 25 & 27

Problems 39, 41, 42, 45, 47, 49

Problems 51, 53, 69,71

Problems continue of 71, & 73