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Calculus, EXT2 C1 EQ-Bank 1

Using partial fractions, show that

`int_0^(1/2) 8/(1-x^4)\ dx = log_e 9-4 tan^(−1) (1/2)`  (3 marks)

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`text(See Worked Solutions)`

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`text(Using partial fractions:)`

`8/(1-x^4) = A/(1-x^2) + B/(1 + x^2)`

`A(1 + x^2) + B(1-x^2)` `= 8`
`A + B + (A-B)x^2` `= 8`
`A + B` ` = 8\ \ …\ (1)`
`A-B` ` = 0\ \ …\ (2)`

 
`A = 4, \ B = 4`

 

`4/(1-x^2) = A/(1-x) + B/(1 + x)`

`A(1 + x) + B(1-x)` `= 4`
`A + B + (A-B)x` `= 4`
`A + B` `= 4\ \ …\ (1)`
`A-B` `= 0\ \ …\ (2)`

 
`A = 2, B = 2`
 

`int_0^(1/2) 8/(1-x^4)\ dx` `= int_0^(1/2) 2/(1-x) + 2/(1 + x) + 4/(1 + x^2)\ dx`
  `= [−2ln |1-x| + 2ln |1 + x| + 4tan^(−1)x]_0^(1/2)`
  `= [2ln |(1 + x)/(1-x)| + 4tan^(−1)x]_0^(1/2)`
  `= 2ln |(1 1/2)/(1/2)| + 4tan^(−1)(1/2)-(2ln1 + 4tan^(−1) 0)`
  `= 2ln3 + 4tan^(−1)(1/2)`
  `= ln9 + 4tan^(−1)(1/2)`

Calculus, EXT2 C1 2008 HSC 1e

It can be shown that

`qquad (8(1-x))/((2-x^2)(2-2x+x^2)) = (4-2x)/(2-2x+x^2)-(2x)/(2-x^2)`    (do NOT prove this)
 

Use this result to evaluate  `int_0^1 (8(1-x))/((2-x^2)(2-2x+x^2))\ dx`   (4 marks)

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`pi/2`

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`int_0^1 (8(1-x))/((2-x^2)(2-2x+x^2))\ dx`

`=int_0^1 (4-2x)/(2-2x+x^2)\ dx-int_0^1 (2x)/(2-x^2)\ dx`

`=int_0^1 (2-(2x-2))/(2-2x+x^2)\ dx + [log_e|2-x^2|]_0^1`

`=int_0^1 2/(1+(1-x)^2)-int_0^1 (2x-2)/(2-2x+x^2) + [log_e|2-x^2|]_0^1`

`=[-2tan^(-1)(1-x)]_0^1-[log_e(2-2x+x^2)]_0^1 + [log_e|2-x^2|]_0^1`

`=-2tan^(-1)0+2tan^(-1) 1-(log_e1-log_e2) + (log_e 1-log_e2)`

`=0+2 xx pi/4 + log_e2-log_e 2`

`=pi/2`

Calculus, EXT2 C1 2017 HSC 14a

It is given that  `x^4 + 4 = (x^2 + 2x + 2) (x^2-2x + 2)`.

  1. Find `A` and `B` so that  `16/(x^4 + 4) = (A + 2x)/(x^2 + 2x + 2) + (B-2x)/(x^2-2x + 2)`.  (1 mark)

  2. Hence, or otherwise, show that for any real number `m`
     
         `int_0^m 16/(x^4 + 4)\ dx = ln ((m^2 + 2m + 2)/(m^2-2m + 2)) + 2 tan^(-1) (m + 1) + 2 tan^(-1) (m-1)`.  (2 marks)

  3. Find the limiting value as  `m -> oo`  of
     
         `int_0^m 16/(x^4 + 4)\ dx`.  (1 mark)

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  1. `A = 4 and B = 4`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `2 pi`
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i.  `16/(x^4 + 4) = (A + 2x)/(x^2 + 2x + 2) + (B-2x)/(x^2-2x + 2)`

`text(Multiply each side by)\ \ x^4 + 4`

`16 = (A + 2x) (x^2-2x + 2) + (B-2x) (x^2 + 2x + 2)`

`text(When)\ \ x = 0,`

`2A + 2B` `=16`
`A + B` `= 8\ text{… (1)}`

 
`text(When)\ \ x = 1`

`(A + 2) (1) + (B-2) (5)` `=16`
`A + 2 + 5B-10` `=16`
`A + 5B` `= 24\ text{… (2)}`

 

`text{Subtract  (2) – (1)}`

`4B=16\ \ =>\ \ B=4`

`A=4`
 

ii.  `int_0^m 16/(x^4 + 4)`

`= int_0^m (4 + 2x)/(x^2 + 2x + 2) dx + int_0^m (4-2x)/(x^2-2x + 2) dx`

`= int_0^m (2x + 2)/(x^2 + 2x + 2) + 2/(x^2 + 2x + 2) dx + int_0^m 2/(x^2-2x + 2)-(2x-2)/(x^2-2x + 2) dx`

`= [ln(x^2 + 2x + 2)]_0^m + int_0^m 2/(1 + (x + 1)^2) dx + int_0^m 2/(1 + (x-1)^2) dx-[ln (x^2-2x + 2)]_0^m`

`= [ln(m^2 + 2m + 2)-ln 2] + [2 tan^(-1) (x + 1)]_0^m + [2 tan^(-1) (x-1)]_0^m-[ln (m^2-2m + 2)-ln 2]`

`= ln ((m^2 + 2m + 2)/(m^2-2m + 2)) + 2 [tan^(-1) (m + 1)-tan^(-1) (1)] + 2 [tan^(-1) (m-1)-tan^(-1) (1)]`

`= ln ((m^2 + 2m + 2)/(m^2-2m + 2)) + 2 tan^(-1) (m + 1)-2 · pi/4 + 2 tan ^(-1) (m-1) + 2 · pi/4`

`= ln ((m^2 + 2m + 2)/(m^2-2m + 2)) + 2 tan^(-1) (m + 1) + 2 tan^(-1) (m-1)`

 

iii.  `I = int_0^m 16/(x^4 + 4) = ln ((1 + 2/m + 2/m^2)/(1-2/m + 2/m^2)) + 2 tan^(-1) (m + 1) + 2 tan^(-1) (m-1)`

`lim_(m -> oo) I` `= ln\ 1 + 2 · pi/2 + 2 · pi/2`
  `= 0 + pi + pi`
  `= 2 pi`