## 5.8 - 5.10 &  6.1- 6.7

(First exam to be May 27th (4th week of class))

PS... If you have specific problems that  you want on here, type them up then e-mail them to me

other wise I chose ones that I thought where of interest for some reason....

most likely it was the algebra in the problem.  Any input to this site is appreciated!

### Chapter 5 section 8

Sec 5.8 pg 386 # 47

g(x) = (arcsin 3x)/ x

First rearrange the problem

to make it easier to work with.

g(x) = (1/x) * (arcsin 3x)

Things needed to do the problem.

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

let u = (3x) then u' = 3

therefore:

(d/dx)[arcsin u] = [3 / (sqrt (1 - (3x)^2))]

or  simplified:  [3 / (sqrt (1 - (9 x^2))]

Now diff of eqn:

g ' (x) = [ (d/dx)(1/x) *(arcsin 3x)] + [ (d/dx)(arcsin 3x) *(1/x)]

g ' (x) = [ (-1/(x^2) *(arcsin 3x)] + [3 / (sqrt (1 - (9 x^2)) *(1/x)]

g ' (x) = [ (-arcsin 3x)/(x^2) ] + [3 / (x*(sqrt (1 - (9 x^2)))]

Rearrange to make it easier to work.

g ' (x) = [3 / (sqrt (1 - (9 x^2))] + [ (-arcsin 3x)/(x^2) ]

or  simplified further.

g ' (x) = [3 / (sqrt (1 - (9 x^2))] - [ (arcsin 3x)/(x^2) ]

Get a common denominator:

[(x^2)*(sqrt (1 - (9x^2)))]

Common denominator of 1st part:

[3 / (x*(sqrt (1 - (9 x^2)))] * [x / x]

= [(3 * x )/(x ^2* (sqrt (1 - (9 x^2)))]

Common denominator of other part

[(arcsin 3x)/(x^2) ]* [(sqrt (1 - (9 x^2))) /

(sqrt (1 - (9 x^2)))]

= [(sqrt (1 - (9 x^2)))*(arcsin 3x) /

(x^2 * (sqrt (1 - (9 x^2))))

Then combine and simplify

Final solution:

g ' (x) = [3 x - (sqrt (1 - (9x^2))) * (-arcsin 3x)]  / [(x^2)*(sqrt (1 - (9x^2)))

 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] Simplify the problem by combining (1/2)*(1/2). y = (1/4)*[(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 (x^2 + 1) 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. y = [8 * arcsin( x / 4)] - [(x*(16 - x^2)^(1/2)))/2  ] 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))] Now diff of eqn: 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))) ] Now you need get a common denominator of (2 (sqrt (16 - ( x ^2))....  I will skip showing that step. Final solution: y ' (x) = ( x ^2) /(sqrt (16 - ( x ^2))

Page 387 # 73

 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

### Chapter 5 section 9

 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. Sometimes it help to move the dx to Numerator for ease of understanding the problem. ∫ (1*dx / ((x)^(1/2)*(1-x)^(1/2))) ∫ (dx / ((x)^(1/2)*(1-x)^(1/2))) 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 Now re-substitute u and a. = 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 5.9If you want others added, e-mail the problems to me and I will post them.

### Chapter 5 Section 10

 I got too busy to type the problems so you will have to suffer with reading my writing.   The good thing is that all assigned problems are shown instead of only a selected few.

#### Problems 25 & 27

 Things needed to do the problem.

 Things needed to do the problem.

 Things needed to do the problem.

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