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Calculus, MET2 2022 VCAA 11 MC

If `\frac{d}{d x}(x \cdot \sin(x))=\sin (x)+x \cdot \cos(x)`, then `\frac{1}{k} \int x \ cos(x)dx` is equal to

  1. `k\left(x \cdot \sin (x)-\int \sin (x) d x\right)+c`
  2. `\frac{1}{k} x \cdot \sin (x)-\int \sin (x) d x+c`
  3. `\frac{1}{k}\left(x \cdot \sin (x)-\int \sin (x) d x\right)+c`
  4. `\frac{1}{k}(x \cdot \sin (x)-\sin (x))+c`
  5. `\frac{1}{k}\left(\int x \cdot \sin (x) d x-\int \sin (x) d x\right)+c`
Show Answers Only

`C`

Show Worked Solution

Given `\frac{d}{d x}(x \cdot \sin(x))=\sin (x)+x \cdot \cos(x)`, then

`x\cdot \cos (x)` `=\frac{d}{d x}(x\cdot \sin (x))-\sin (x)`  
`\frac{1}{k} \int x \cos (x) d x` `= \frac{1}{k}\left(\int \frac{d}{d x}(x \cdot \sin (x)) d x-\int \sin (x) d x\right)`  
  `= \frac{1}{k}\left(x \cdot \sin (x)-\int \sin (x)d x\right)+c`  

 
`=> C`

Calculus, MET1 2014 ADV 13a

  1. Differentiate  `3 + sin 2x`.    (1 mark)
  2. Hence, or otherwise, find  `int (cos2x)/(3 + sin 2x)\ dx`.    (2 marks)
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  1. `2 cos 2x`
  2. `ln (3 + sin 2x)^(1/2) + C`
Show Worked Solution
a. `y` `= 3 + sin 2x`
  `dy/dx` `= 2 cos 2x`

 

b. `int (cos 2x)/(3 + sin 2x)\ dx`
  `= 1/2 int (2 cos 2x)/(3 + sin 2x)\ dx`
  `= 1/2  ln (3 + sin 2x) + C\ \ \ \ \ text{(from part (i))}`

Calculus, MET1 2008 ADV 3b

  1. Differentiate  `log_e(cos x)` with respect to `x`.  (2 marks)
  2. Hence, or otherwise, evaluate  `int_0^(pi/4) tan x\ dx`.  (2 marks)
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  1. `−tan x`
  2. `−log_e(1/sqrt2)\  text{or  0.35  (2 d.p.)}`
Show Worked Solution
a.   `y` `= log_e(cos x)`
  `(dy)/(dx)` `= (−sin x)/(cos x)`
    `= −tan x`

 

b.   `int_0^(pi/4) tan x\ dx`

`= −[log_e(cos x)]_0^(pi/4)`

`= −[log_e(cos(pi/4)) – log_e(cos 0)]`

`= −[log_e(1/sqrt2) – log_e 1]`

`= −[log_e(1/sqrt2) – 0]`

`= −log_e(1/sqrt2)`

`= 0.346…`

`= 0.35\ \ (text(2 d.p.))`

Calculus, MET1 2007 VCAA 7

If  `f(x) = xcos(3x)`,  then  `f′(x) = cos(3x) - 3xsin(3x)`.

Use this fact to find an antiderivative of  `xsin(3x)`.  (3 marks)

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`- 1/3xcos(3x) + 1/9sin(3x)`

Show Worked Solution

`text(If)\ \ f′(x) = g(x) \ -> \ int g(x)\ dx = f(x)`

♦ Mean mark 36%.
MARKER’S COMMENT: This standard and familiar calculus problem was surprisingly poorly answered.
`int (cos(3x) -3xsin(3x))\ dx` `= xcos(3x)`
`1/3 sin(3x) – 3 int xsin(3x)\ dx` `= xcos(3x)`
`-3 int xsin(3x)\ dx` `= xcos(3x) – 1/3sin(3x)`
`:. int xsin(3x)\ dx` `= 1/9sin(3x) – 1/3xcos(3x)`

Calculus, MET1 2012 VCAA 9

  1. Let  `f: R -> R,\ \ f(x) = x sin (x)`.
  2. Find  `f prime (x)`.  (1 mark)
  3. Use the result of part a. to find the value of  `int_(pi/6)^(pi/2) x cos (x)\ dx`  in the form  `a pi + b`.  (3 marks)
Show Answers Only
  1. `x cos (x) + sin (x)`
  2. `(5 pi)/12 – sqrt 3/2`
Show Worked Solution

a.   `f(x)=x sin(x)`

`text(Using Product Rule:)`

`(gh)′` `= g′ h + g h′`
`:. f prime (x)` `= x cos (x) + sin (x)`

 

b.   `text(Integrating)\ \ f′(x)\ \ text{from part (a),}`

`int (x cos (x) + sin (x))\ dx` `= x sin (x)`
`int (x cos x)\ dx – cos (x)` `= x sin (x)`
`:. int (x cos x)\ dx` `= x sin (x) + cos x`

 

`text(Evaluate definite integral:)`

`int_(pi/6)^(pi/2) (x cos x) dx`

`= [x sin (x) + cos (x)]_(pi/6)^(pi/2)`

`= (pi/2 sin (pi/2) + cos (pi/2)) – (pi/6 sin (pi/6) + cos (pi/6))`

`= (pi/2 + 0) – (pi/6 xx 1/2 + sqrt 3/2)`

`= (5 pi)/12 – sqrt 3/2`