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Calculus, EXT2 C1 2023 HSC 13a

Find \({\displaystyle \int \frac{1-x}{\sqrt{5-4 x-x^2}}\ dx}\).  (3 marks)

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\(I=\sqrt{5-4x-x^2} + 3 \sin^{-1} \big{(} \dfrac{x+2}{3} \big{)} + c \)

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\(I\) \(={\displaystyle \int \frac{1-x}{\sqrt{5-4 x-x^2}}\ dx}\)  
  \(= \dfrac{1}{2} {\displaystyle \int \frac{2-2x}{\sqrt{5-4 x-x^2}}\ dx}\)  
  \(=\dfrac{1}{2} {\displaystyle \int \frac{-4-2x}{\sqrt{5-4 x-x^2}}\ dx} + \dfrac{1}{2} {\displaystyle \int \frac{6}{\sqrt{5-4 x-x^2}}\ dx}\)  

 
\(\text{Let}\ \ u=5-4x-x^2 \)

\( \dfrac{du}{dx}=-4-2x\ \ \Rightarrow\ \ du=-4-2x\ dx \)

\(I\) \(= \dfrac{1}{2} {\displaystyle \int u^{-\frac{1}{2}}\ du} + {\displaystyle 3 \int \frac{1}{\sqrt{9-(x+2)^2}}\ dx}\)  
  \(= u^{\frac{1}{2}} + 3 \sin^{-1} \big{(} \dfrac{x+2}{3} \big{)} + c \)  
  \(=  \sqrt{5-4x-x^2}+ 3 \sin^{-1} \big{(} \dfrac{x+2}{3} \big{)} + c \)  

Complex Numbers, EXT2 N1 2020 HSC 13d

  1. Show that for any integer `n`,  `e^(i n theta) + e^(-i n theta) = 2 cos (n theta)`.   (1 mark)

  2. By expanding  `(e^(i theta) + e^(-i theta))^4`  show that
     
       `cos^4 theta = frac{1}{8} ( cos (4 theta) + 4 cos (2 theta) + 3 )`.   (3 marks)

  3. Hence, or otherwise, find  `int_0^(frac{pi}{2}) cos^4 theta\ d theta`.   (2 marks)

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  1. `text{See Worked Solution}`
  2. `text{See Worked Solution}`
  3. `text{See Worked Solution}`
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i.      `e^(i n theta) + e^(-i n theta)` `= cos(n theta) + i sin(n theta) + cos(-n theta) + i sin(-n theta)`
    `= cos(n theta) + i sin(n theta) + cos(n theta) – i sin(n theta)`
    `= 2 cos (n theta)`

 

ii.    `(e^{i theta} + e^{-i theta})^4` `= (2 cos theta)^4`
    `= 16 cos^4 theta`

 
`text{Expand} \ (e^{i theta} + e^{-i theta})^4 :`

`e^(i 4 theta) + 4 e^(i 3 theta) e^(-i theta) + 6 e^(i 2 theta) e^(-i 2 theta) + 4 e^(i theta) e^(-i 3 theta) + e^(-i 4 theta)`

`= e^(i 4 theta) + 4e^(i 2 theta) + 6 + 4^(-i 2 theta) + e^(-i 4 theta)`

`= e^(i 4 theta) + e^(i 4 theta) + 4 (e^{i 2 theta} + e^{-i 2 theta}) + 6`

`= 2 cos (4 theta) + 8 cos (2 theta) + 6`
 

`therefore \ 16 cos^4 theta` `= 2 cos (4 theta) + 8 cos (2 theta) + 6`
`cos^4 theta` `= frac{1}{8} cos(4 theta) + 1/2 cos(2 theta) + 3/8`
`cos^4 theta` `= frac{1}{8} (cos(4 theta) + 4 cos(2 theta) + 3)`

 

iii.    `int_0^(frac{pi}{2}) cos^4 theta\ d theta` `= frac{1}{8} int_0^(frac{pi}{2}) cos(4 theta) + 4 cos(2 theta) + 3\ d theta`
    `= frac{1}{8} [ frac{1}{4} sin(4 theta) + 2 sin (2 theta) + 3 theta ]_0^(frac{pi}{2}`
    `= frac{1}{8} [( frac{1}{4} sin (2 pi) + 2 sin pi  + frac{3 pi}{2}) – 0 ]`
    `= frac{1}{8} ( frac{3 pi}{2})`
    `= frac{3 pi}{16}`

Calculus, EXT2 C1 2019 HSC 2 MC

Which of the following is a primitive of  `(sin x)/(cos^3 x)`?

  1. `1/2 sec^2 x`
  2. `-1/2 sec^2 x`
  3. `1/4 sec^4 x`
  4. `-1/4 sec^4 x`
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`A`

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`text(Solution 1)`

`int(sin x)/(cos^3 x)` `= int sinx cos^(-3) x\ dx`
  `=1/2 cos^(-2)x + C`
  `= 1/2 sec^2 x + C`

 
`text(Solution 2)`

`int(sin x)/(cos^3 x)` `= int tanx sec^2x\ dx`
  `= 1/2 tan^2 + C_1`
  `= 1/2 (sec^2 x -1) + C_1`
  `=1/2 sec^2 x + C_2`

 
`=>A`

Calculus, EXT2 C1 2015 HSC 14a

  1. Differentiate  `sin^(n - 1) theta cos theta`, expressing the result in terms of  `sin theta`  only.  (2 marks)

  2. Hence, or otherwise, deduce that
     
         `int_0^(pi/2) sin^n theta\ d theta = ((n-1))/n int_0^(pi/2) sin^(n - 2) theta\ d theta`,  for `n>1.`  (2 marks) 

  3. Find  `int_0^(pi/2) sin^4 theta\ d theta.`  (1 mark)

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  1. `text(See Worked Solutions)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `(3 pi)/16`
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i.   `d/(d theta) (sin^(n – 1) theta cos theta)`

`=(n – 1) sin^(n – 2) theta cos theta cos theta + sin^(n – 1) theta xx (-sin theta)`

`=(n – 1) sin^(n – 2) theta cos^2 theta – sin^n theta`

`=(n – 1) sin^(n – 2) theta (1 – sin^2 theta) – sin^n theta`

`=(n – 1) sin^(n – 2) theta – (n – 1) sin^n theta – sin^n theta`

`=(n – 1) sin^(n – 2) theta – n sin^n theta`

 

ii.   `text{From part (i)}`

`n sin^n theta = (n – 1) sin^(n – 2) theta – d/(d theta) (sin^(n – 1) theta cos theta)`

`:. int_0^(pi/2) sin^n theta\ d theta`

`=1/n int_0^(pi/2) ((n – 1) sin^(n – 2) theta – d/(d theta) (sin^(n – 1) theta cos theta)) d theta`

`=1/n int_0^(pi/2) (n – 1) sin^(n – 2) theta\ d theta – 1/n int_0^(pi/2) d/(d theta) (sin^(n – 1) theta cos theta)\ d theta`

`= 1/n int_0^(pi/2) (n – 1) sin^(n – 2) theta\ d theta – 1/n int_0^(pi/2) d/(d theta) (sin^(n – 1) theta cos theta)\ d theta`

`= (n – 1)/n int_0^(pi/2) sin^(n – 2) theta\ d theta – 1/n [sin^(n – 1) theta cos theta]_0^(pi/2)`

`= (n – 1)/n int_0^(pi/2) sin^(n – 2) theta\ d theta – 1/n (0 – 0)`

`= (n – 1)/n int_0^(pi/2) sin^(n – 2) theta\ d theta,\ \ \ \ (n>1)`

 

iii.   `int_0^(pi/2) sin^4 theta\ d theta` `= 3/4 int_0^(pi/2) sin^2 theta\ d theta`
    `= 3/4 xx [(2-1)/2 int_0^(pi/2) d theta]`
    `= 3/8 xx [theta]_0^(pi/2)` 
    `= (3 pi)/16`

Calculus, EXT2 C1 2011 HSC 7b

Let   `I = int_1^3 (cos^2(pi/8 x))/(x(4-x))\ dx.`

  1. Use the substitution  `u = 4-x`  to show that
     
          `I = int_1^3 (sin^2(pi/8 u))/(u(4-u))\ du.`  (2 marks)

     

  2. Hence, find the value of  `I`.  (3 marks)

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  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `1/4 log_e 3`
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i.   `text(Let)\ \ u = 4-x,\ \ du = -dx,\ \ x = 4-u`

`text(When)\ \ x = 1,\ \ u = 3`

`text(When)\ \ x = 3,\ \ u = 1`

`I` `=int_1^3 (cos^2(pi/8 x))/(x(4-x))\ dx`
  `=int_3^1 (cos^2 (pi/8(4-u)))/((4-u)u) (-du)`
  `=-int_3^1 (cos^2(pi/2-pi/8 u))/((4-u)u)\ du`
  `=int_1^3 (sin^2(pi/8 u))/(u (4-u))\ du`

 

ii.   `text{S}text{ince}\ \ int_1^3 (cos^2(pi/8 x))/(x(4-x))\ dx = int_1^3 (sin^2(pi/8 x))/(x(4-x))\ dx`

 

`text(We can add the integrals such that)`

`2I` `=int_1^3 (cos^2(pi/8 x))/(x(4-x)) dx + int_1^3 (sin^2(pi/8 x))/(x(4-x)) dx`
  `=int_1^3 1/(x(4-x)) dx`

 

`text(Using partial fractions:)`

♦ Mean mark 36%.
`1/(x(4-x))` `=A/x+B/(4-x)`
`1` `=A(4-x)+Bx`

 
`text(When)\ \ x=0,\ \ A=1/4`

`text(When)\ \ x=0,\ \ B=1/4`

`2I` `=1/4 int_1^3 (1/x + 1/(4-x))\ dx`
  `=1/4 [log_e x-log_e (4-x)]_1^3`
  `=1/4 [log_e 3-log_e 1-(log_e 1-log_e 3)]`
  `=1/2 log_e 3`
`:.I` `=1/4 log_e 3`

Calculus, EXT2 C1 2011 HSC 1d

Find  `int cos^3 theta\ d theta`  (3 marks)

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`sin theta-(sin^3 theta)/3 + c`

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`int cos^3 theta\ d theta` `= int cos^2 theta cos theta\ d theta`
  `= int (1-sin^2 theta) cos theta\ d theta`
  `= int (cos theta-sin^2 theta cos theta) d theta`
  `= sin theta-(sin^3 theta)/3 + c`