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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`
Show Worked Solution

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 2015 HSC 11f

  1. Show that  
     
    `cot theta + text(cosec)\ theta = cot(theta/2).`   (2 marks)

  2. Hence, or otherwise, find
     
         `int (cot theta + text(cosec)\ theta)\ d theta.`   (1 mark)

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  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `2 ln\ |sin\ theta/2| + c`
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i.   `cot theta + text(cosec)\ theta ­=` `(cos theta)/(sin theta) + 1/(sin theta)`
`­=` `(1 + cos theta)/(sin theta)`
`­=` `(1 + 2 cos^2 (theta/2) – 1)/(2 sin (theta/2) cos (theta/2))`
`­=` `(2 cos^2(theta/2))/(2 sin (theta/2) cos (theta/2))`
`­=` `(cos (theta/2))/(sin (theta/2))`
`­=` `cot (theta/2)`

 

COMMENT: Part (ii) mean mark 51%.
ii.   `int (cot theta + text(cosec)\ theta)\ d theta ­=` `int cot (theta/2)\ d theta`
`­=` `int (cos (theta/2))/(sin (theta/2))\ d theta`
`­=` `2 ln\ |sin\ theta/2| + c`

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 2013 HSC 14c

  1. Given a positve integer `n`, show that
     
         `sec^(2n) theta = sum_(k = 0)^n ((n),(k)) tan^(2k) theta.`  (1 mark)

  2. Hence, by writing  `sec^8 theta`  as  `sec^6 theta\ sec^2 theta` find,
     
         `int sec^8 theta\ d theta.`  (2 marks)

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  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `tan theta + tan^3 theta + 3/5 tan^5 theta + 1/7 tan^7 theta + C`
Show Worked Solution
i.     `sec^2 theta` `= 1 + tan^2 theta`
`sec^(2n) theta` `= (1 + tan^2 theta)^n`
  `= ((n), (0)) + ((n), (1)) tan^2 theta + … + ((n), (k)) tan^(2k) theta +`
  ` … + ((n), (n)) tan^(2n) theta`
♦♦ Mean mark 30%.

MARKER’S COMMENT: Note that  `sec^(2n) theta“ ≠ 1+tan^(2n) theta`. A very common error!

`:.sec^(2n) theta= sum_(k = 0)^n ((n), (k)) tan^(2k) theta\ \ \ \ text(… as required)`

 

♦ Mean mark 38%.

ii.  `int sec^8 theta\ d theta` `= int sec^6 theta sec^2 theta\ d theta`
  `= int (1 + tan^2 theta)^3 sec^2 theta\ d theta`
  `= int [1 + ((3), (1)) tan^2 theta + ((3), (2)) tan^4 theta`
  `+ ((3), (3)) tan^6 theta] sec^2 theta\ d theta`
  `=int(1 + 3 tan^2 theta+3 tan^4 theta+tan^6 theta)sec^2 theta\ d theta`
  `= tan theta + tan^3 theta + 3/5 tan^5 theta + 1/7 tan^7 theta + c`