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Calculus, 2ADV C4 EQ-Bank 26

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

Show Worked Solution

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 2023 MET1 5

  1. Evaluate  \(\displaystyle \int_{0}^{\frac{\pi}{3}} \sin(x)\,dx\).   (1 mark)
  2. Hence, or otherwise, find all values of  \(k\)  such that  \(\displaystyle \int_{0}^{\frac{\pi}{3}} \sin(x)\,dx=\displaystyle \int_{k}^{\frac{\pi}{2}} \cos(x)\,dx\), where  \(-3\pi<k<2\pi\).   (3 marks)
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a.    \(\dfrac{1}{2}\)

b.    \(k=\dfrac{-11\pi}{6},\ \dfrac{-7\pi}{6},\ \dfrac{\pi}{6},\ \dfrac{5\pi}{6}\)

Show Worked Solution
a.     \(\displaystyle \int_{0}^{\frac{\pi}{3}} \sin(x)\,dx\) \(=\left[-\cos x\right]_0^\frac{\pi}{3}\)
    \(=-\cos\dfrac{\pi}{3}+\cos 0\)
    \(=-\dfrac{1}{2}+1\)
    \(=\dfrac{1}{2}\)

 

b.     \(\displaystyle \int_{k}^{\frac{\pi}{2}} \cos(x)\,dx\) \(=\left[\sin x\right]_k^\frac{\pi}{2}\)
    \(=\sin\bigg(\dfrac{\pi}{2}\bigg)-\sin (k)\)
    \(=1-\sin (k)\)

 
\(\text{Using part (a):}\)

\(1-\sin (k)\) \(=\dfrac{1}{2}\)
\(\sin (k)\) \(=\dfrac{1}{2}\)
\(\therefore\ k\) \(=\dfrac{-11\pi}{6},\ \dfrac{-7\pi}{6},\ \dfrac{\pi}{6},\ \dfrac{5\pi}{6}\)

Calculus, 2ADV C4 2008 HSC 5a

The gradient of a curve is given by  `dy/dx = 1-6 sin 3x`. The curve passes through the point  `(0, 7)`.

What is the equation of the curve?   (3 marks)

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`y = x + 2 cos 3x + 5`

Show Worked Solution
`dy/dx` `= 1-6 sin 3x`
`y` `= int 1-6 sin 3x\ dx`
  `= x + 2 cos 3x + c`

  
`text(Passes through)\ (0,7):`

`=> 0 + 2 cos 0 + c` `= 7`
`2 + c` `= 7`
`c` `= 5`

  
`:.\ text(Equation is)\ \ \ y = x + 2 cos 3x + 5`

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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a.    `-tan x`

b.    `-log_e (1/sqrt2)\ \ text(or)\ \ 0.35\ \ text{(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, 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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a.    `2 cos 2x`

b.    `1/2  ln (3 + sin 2x) + C\ \ \ \ \ text{(from part (a))}`

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 (a))}`

Calculus, 2ADV C4 2010 HSC 5b

  1. Prove that  `sec^2 x + secx\ tanx = (1 + sinx)/(cos^2x)`.   (1 mark)

  2. Hence prove that  `sec^2 x + secx\ tanx = 1/(1-sinx)`.   (1 mark)

  3. Hence, use the identity  `int sec(ax)tan(ax)\ dx=1/a sec(ax)`  to find the exact value of

     

          `int_0^(pi/4) 1/(1-sinx)\ dx`.   (2 marks)

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a.    `text(Proof)  text{(See Worked Solutions)}`

b.    `text(Proof)  text{(See Worked Solutions)}`

c.    `sqrt2`

Show Worked Solution

a.    `text(Need to prove)`

`sec^2x + secxtanx = (1 + sinx)/(cos^2x)`
 

`text(LHS)` `=sec^2x + secx tanx`
  `=1/(cos^2x) + 1/(cosx) xx (sinx)/cosx`
  `=1/(cos^2x) + (sinx)/(cos^2x)`
  `=(1 + sinx)/(cos^2x)= text(RHS)\ \ \ \ text(… as required)`

  
b.    `text(Need to prove)`

♦♦ Mean mark 31%.
`sec^2x + secx tanx` `= 1/(1-sinx)`
`text(i.e.)\ \ (1 + sinx)/(cos^2x)` `= 1/(1-sin x)\ \ \ \ \ text{(part (a))}`
`text(LHS)` `= (1 + sinx)/(cos^2x)`
  `=(1 + sin x)/(1-sin^2x)`
  `=(1 + sinx)/((1-sinx)(1 + sinx)`
  `=1/(1-sinx)\ \ \ \ text(… as required)`

  
c.    `int_0^(pi/4) 1/(1-sinx)\ dx`

♦ Mean mark 37%.

`= int_0^(pi/4) (sec^2x + secx\ tanx)\ dx`

`= [tanx + secx]_0^(pi/4)`

`= [(tan(pi/4) + sec(pi/4))-(tan0 + sec0)]`

`= [(1 + 1/(cos(pi/4)))-(0 + 1/(cos0))]`

`= 1 + sqrt2-1= sqrt2`