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Calculus, 2ADV C4 EQ-Bank 3

  1. Differentiate  \(y=x\, \sin (2 x)\).   (2 marks)

  2. Hence, or otherwise, find \(\displaystyle \int x\, \cos (2 x)\, d x\).  (2 marks)

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a.  \(\dfrac{d y}{d x}=\sin (2 x)+2 x\, \cos (2 x)\)

b.   \(\dfrac{1}{2} x\, \sin (2 x)+\dfrac{1}{4} \cos (2 x)+c\)

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a.     \(y=x\, \sin (2 x)\)

\(\dfrac{d y}{d x}=\sin (2 x)+2 x\, \cos (2 x)\)
 

b.    \(\text {Using part a.}\)

\(\displaystyle\int \sin (2 x)+2 x\, \cos (2 x)\, d x=x\, \sin (2 x)+c\)

\(\displaystyle\int \sin (2 x)\, d x+2 \int x\, \cos (2 x)\, d x=x\, \sin (2 x)+c\)

\(2 \displaystyle \int x\, \cos (2 x)\, d x\) \(=x\, \sin (2 x)-\displaystyle \int \sin (2 x)\, d x+c\)
  \(=x\, \sin (2 x)+\dfrac{1}{2} \cos (2 x)+c\)
\(\displaystyle\int x\, \cos (2 x)\, d x\) \(=\dfrac{1}{2} x\, \sin (2 x)+\dfrac{1}{4} \cos (2 x)+c\)

Calculus, 2ADV C4 2024 HSC 27

  1. Find the derivative of  \(x^{2}\tan\,x\)   (2 marks)
  2. Hence, find \(\displaystyle \int (x\,\tan\,x+1)^2\ dx\)   (3 marks)
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a.   \(\dfrac{dy}{dx}=2x\,\tan\,x + x^2\,\sec^{2}x\)

b.   \(x^{2}\tan^{2}x-\dfrac{x^{3}}{3}+x+C\)

Show Worked Solution

a.  \(y=x^{2}\tan\,x\)

\(\text{By product rule:}\)

\(\dfrac{dy}{dx}=2x\,\tan\,x + x^2 \sec^{2}x\)
 

b.   \(\displaystyle \int (x\,\tan\,x+1)^{2}\,dx\)

\[=\int x^2\tan^{2}x + 2x\,\tan\,x +1\ dx\]

\[=\int x^{2}(\sec^{2}x-1)+2x\,\tan\,x +1\,dx\]

\[=\int 2x\,\tan\,x + x^{2}\sec^{2}x-x^{2}+1\,dx\]

\[=x^{2}\tan\,x-\dfrac{x^{3}}{3}+x+C\]

♦♦ Mean mark (b) 34%.

Calculus, 2ADV C4 2008 HSC 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\ \ text{(2 d.p.)}`
Show Worked Solution
i.    `y` `= log_e (cos x)`
  `dy/dx` `= (- sin x)/(cos x)`
    `= – tan x`

 

ii.    `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, 2ADV C4 2014 HSC 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
i. `y` `= 3 + sin 2x`
  `dy/dx` `= 2 cos 2x`

 

ii. `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))}`