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Calculus, EXT2 C1 2025 HSC 14a

Let  \(\displaystyle I_n=\large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2 n} \theta \, d \theta\)  for integers  \(n \geq 0\).

  1. Show that  \(I_n=\dfrac{1}{2 n-1}-I_{n-1}\) for \(n>0\),  given that  \(\dfrac{d}{d \theta} \cot \theta=-\operatorname{cosec}^2 \theta\).   (3 marks)

  2. Hence, or otherwise, calculate \(I_2\).   (1 mark)

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i.    \(\text{See Worked Solutions}\)

ii.   \(\dfrac{\pi}{4}-\dfrac{2}{3}\)

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i.    \(\displaystyle I_n=\large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2 n} \theta \, d \theta \ \ \Rightarrow \ \ \)\(I_{n-1}=\displaystyle \large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2n-2} \theta \, d \theta\)

\(\text{Show} \ \ I_n=\dfrac{1}{2n-1}-I_{n-1} \ \ \text{for} \ \ n>0\)

\(\text{Given} \ \ \dfrac{d}{d \theta} \cot \theta=-\operatorname{cosec}^2 \theta\ \ldots\ (1)\)

\(I_n\) \(=\displaystyle \large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2 n-2} \theta \cdot \cot ^2 \theta \, d \theta\)
  \(=\displaystyle \large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2 n-2} \theta\left(\operatorname{cosec}^2 \theta-1\right) \, d \theta\)
  \(=\displaystyle \large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2 n-2} \theta \cdot \operatorname{cosec}^2 \theta \, d \theta-\displaystyle \large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2n-2} \theta \, d \theta\)
  \(=-\displaystyle \large{\int}_{\small{\dfrac{\pi}{4}}}^{\small{\dfrac{\pi}{2}}}\)\(\cot ^{2 n-2} \theta \cdot \dfrac{d}{d \theta}(\cot \theta) d \theta-I_{n-1}\)
  \(=-\left[\dfrac{\cot ^{2 n-1} \theta}{2 n-1}\right]_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}-I_{n-1}\)
  \(=-\left[\dfrac{\cot ^{2 n-1} (\frac{\pi}{2})}{2 n-1}-\dfrac{\cot ^{2 n-1} (\frac{\pi}{4})}{2 n-1}\right]-I_{n-1}\)
  \(=-\left[0-\dfrac{1}{2 n-1}\right]-I_{n-1}\)
  \(=\dfrac{1}{2n-1}-I_{n-1}\)

 

ii.     \(I_2\) \(=\dfrac{1}{3}-I_1\)
    \(=\dfrac{1}{3}-\left[ \dfrac{1}{2-1}-\displaystyle \int_{\tfrac{\pi}{2}}^{\tfrac{\pi}{2}} 1 \, d\theta \right]\)
    \(=\dfrac{1}{3}-\left[1-\left(\dfrac{\pi}{2}-\dfrac{\pi}{4}\right)\right]\)
    \(=\dfrac{\pi}{4}-\dfrac{2}{3}\)

Calculus, EXT2 C1 2023 HSC 15a

  1. Let  \(J_n= {\displaystyle \int_0^{\frac{\pi}{2}}} \sin ^n \theta \ d \theta\)  where \(n \geq 0\) is an integer. 
  2. Show that  \(J_n=\dfrac{n-1}{n} J_{n-2}\)  for all integers \(n \geq 2\).  (3 marks)

  3. Let  \(I_n={\displaystyle \int_0^1 x^n(1-x)^n}\, dx \)  where \( n \)  is a positive integer.
  4. By using the substitution  \(x=\sin ^2 \theta\), or otherwise,
  5. show that  \( I_n=\dfrac{1}{2^{2 n}} {\displaystyle \int_0^{\frac{\pi}{2}}} \sin ^{2 n+1} \theta \ d \theta \).  (4 marks)

  6. Hence, or otherwise, show that  \(I_n=\dfrac{n}{4 n+2} I_{n-1}\), for all integers  \(n \geq 1\).   (2 marks)
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  1. \(\text{See Worked Solutions}\)
  2. \(\text{See Worked Solutions}\)
  3. \(\text{See Worked Solutions}\)
Show Worked Solution
i.    \(J_n\) \(= \displaystyle \int_0^{\frac{\pi}{2}} \sin^{n}\, \theta\, d\theta \)  
    \(= \displaystyle \int_0^{\frac{\pi}{2}} \sin^{n-1}\, \theta \cdot \sin\,\theta\, d\theta \)  

 
\(\text{Integrating by parts:} \)

\(u\) \(=\sin^{n-1}\, \theta \) \(u^{′}=(n-1) \cos\, \theta \cdot \sin^{n-2}\,\theta \)
\(v\) \(=-\cos\,\theta \) \(v^{′}=\sin \, \theta\)
\(J_n\) \(= \big{[} -\cos\,\theta \cdot \sin^{n-1}\,\theta \big{]}_0^{\frac{\pi}{2}} + (n-1) \displaystyle \int_0^{\frac{\pi}{2}} \cos^{2} \theta \cdot \sin^{n-2}\theta\, d\theta\)  
  \(=(n-1) \displaystyle \int_0^{\frac{\pi}{2}} (1-\sin^{2} \theta) \sin^{n-2}\theta\, d\theta\)  
  \(=(n-1) \displaystyle \int_0^{\frac{\pi}{2}} \sin^{n-2}\theta-\sin^{n}\theta\, d\theta\)  
  \(=(n-1)J_{n-2}-(n-1)J_n \)  
\(J_n+(n-1)J_n \) \(=(n-1)J_{n-2} \)  
\(J_n(1+n-1) \) \(=(n-1) J_{n-2} \)  
\(J_n\) \(= \dfrac{n-1}{n} J_{n-2} \)  

 

ii.   \(I_n= \displaystyle \int_0^1 x^{n}(1-x)^n\,dx \)

\(\text{Let}\ \ x=\sin^{2}\,\theta \)

\(\dfrac{dx}{d\theta}=2\sin\,\theta\, \cos\,\theta \ \Rightarrow \ dx=2\sin\,\theta \,\cos\,\theta\,d\theta \)

\(\text{When}\ \ x=1, \ \theta=\dfrac{\pi}{2}; \ \ x=0, \ \theta=0 \)

\(I_n\) \(= \displaystyle \int_0^{\frac{\pi}{2}} \sin^{2n}\,\theta(1-\sin^{2}\,\theta)^{n} \cdot 2\sin\,\theta \,\cos\,\theta\,d\theta \)  
  \(= 2 \displaystyle \int_0^{\frac{\pi}{2}} \sin^{2n+1}\,\theta \cdot \cos^{2n+1}\,\theta\,d\theta \)  
  \(=\dfrac{2}{2^{2n+1}} \displaystyle \int_0^{\frac{\pi}{2}} (2\sin\,\theta\,\cos\,\theta)^{2n+1} d\theta \)  
  \(=\dfrac{1}{2^{2n}} \displaystyle \int_0^{\frac{\pi}{2}} \sin^{2n+1}(2\theta)\, d\theta \)  

 
\(\text{Let}\ \ u=2\theta \)

\(\dfrac{du}{d\theta}=2\ \ \Rightarrow\ \ \dfrac{du}{2}=d\theta\)

\(\text{When}\ \ \theta=\dfrac{\pi}{2}, \ u=\pi; \ \theta=0, \ u=0 \)

\(I_n=\dfrac{1}{2^{2n+1}} \displaystyle \int_0^{n} \sin^{2n+1} u\, du\)

 
\(\text{Since}\ \ \sin^{2n+1} u\ \ \text{is symmetrical about}\ \ u=\dfrac{\pi}{2} \)

\(\Rightarrow \displaystyle \int_0^{\pi} \sin^{2n+1} u\,du=2\displaystyle \int_0^{\frac{\pi}{2}} \sin^{2n+1} u\,du \)

\(I_n\) \(= \dfrac{1}{2^{2n+1}} \cdot 2\displaystyle \int_0^{\frac{\pi}{2}} \sin^{2n+1} u\,du \)  
  \(= \dfrac{1}{2^{2n}} \displaystyle \int_0^{\frac{\pi}{2}} \sin^{2n+1} u\,du \)  

 
\(\text{Which can be expressed as:}\)

\(I_n= \dfrac{1}{2^{2n}}  \displaystyle \int_0^{\frac{\pi}{2}} \sin^{2n+1} \theta\, d\theta \)
 

♦♦ Mean mark (ii) 34%.
iii.    \(I_n\) \(=\dfrac{1}{2^{2n}}  \displaystyle \int_0^{\frac{\pi}{2}} \sin^{2n+1} \theta\, d\theta \)
    \(=\dfrac{1}{2^{2n}} J_{2n+1} \ \ \ \text{(by definition)}\)
    \(=\dfrac{1}{2^{2n}} \cdot \underbrace{\dfrac{2n}{2n+1} \cdot J_{2n+1}}_{\text{using part i}}  \)
    \(= \dfrac{n}{2(2n+1)} \cdot \dfrac{2^2}{2^{2n}}\cdot J_{2n-1} \)
    \(= \dfrac{n}{4n+2} \cdot \underbrace{\dfrac{1}{2^{2n-2}}\cdot J_{2n-1}}_{=I_{n-1} \ \ \text{(using part ii)}} \)
    \(=\dfrac{n}{4 n+2} I_{n-1}\)
♦♦ Mean mark (iii) 28%.

Calculus, EXT2 C1 2020 HSC 16b

Let  `I_n = int_0^(frac{pi}{2}) sin^(2n + 1)(2theta)\ d theta, \ n = 0, 1, ...`

  1. Prove that  `I_n = frac{2n}{2n + 1} I_(n-1) , \ n ≥ 1`.  (3 marks)

  2. Deduce that  `I_n = frac{2^(2n)(n!)^2}{(2n +1)!}`.  (3 marks)

Let  `J_n = int_0^1 x^n (1 - x)^n\ dx , \ n = 0, 1, 2,...`

  1. Using the result of part (ii), or otherwise, show that  `J_n = frac{(n!)^2}{(2n + 1)!}`.  (3 marks)

  2. Prove that  `(2^n n!)^2 ≤ (2n + 1)!`.  (2 marks)

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  1. `text{See Worked Solutions}`
  2. `text{See Worked Solutions}`
  3. `text{See Worked Solutions}`
  4. `text{See Worked Solutions}`
Show Worked Solution

i.    `text{Prove} \ \ I_n = frac{2n}{2n + 1} I_(n-1) , \ n ≥ 1`

♦ Mean mark (i) 48%.

`I_n = int_0^(frac{pi}{2}) sin^(2n) (2 theta) * sin (2 theta)\ d theta`

`text{Integrating by parts:}`

`u = sin^(2n) (2 theta)`   `u^(′) = 2n sin^(2n -1) (2 theta)  xx -frac(1)(2) cos (2 theta)`
`v = -frac{1}{2} cos (2 theta)`   `v^(′) = sin 2 theta`

 

`I_n` `= [ sin^(2n) (2 theta) * -frac{1}{2} cos (2 theta)]_0^(frac{pi}{2}) -2n int_0^(frac{pi}{2}) sin^(2n -1) (2 theta) * 2 cos (2 theta) * -frac{1}{2} cos (2 theta)\ d theta`
`I_n` `= 0 + 2n int_0^(frac{pi}{2}) sin^(2n-1) (2 theta) * cos^2 (2 theta)\ d theta`
`I_n` `= 2 n int_0^(frac{pi}{2}) sin^(2n-1) (2 theta) (1 – sin^2 (2 theta))\ d theta`
`I_n` `= 2 n int_0^(frac{pi}{2}) sin^(2n-1) ( 2 theta) – sin^(2n+1) (2 theta)\ d theta`
`I_n` `= 2n (I_(n-1) – I_n)`
`I_n + 2 n  I_n` `= 2 n I_(n-1)`
`I_n (2n + 1)` `= 2 n I_(n-1)`
`therefore I_n` `= frac{2n}{2n +1} I_(n-1)`
♦ Mean mark (ii) 36%.

 

ii.     `I_0` `= int_0^(frac{pi}{2}) sin (2 theta)\ d theta`
    `= [ -frac(1)(2) cos (2 theta) ]_0^(frac{pi}{2}`
    `=( -frac{1}{2} cos pi + frac{1}{2} cos 0 )`
    `= 1`
     
`I_n` `= frac{2n}{2n + 1} I_(n-1)`
`I_(n-1)` `= frac{2(n -1)}{2n -1} I_(n-2)`
  `vdots`
`I_1` `= frac{2}{3} I_0`

 

`I_n` `= frac{2n}{2n + 1} xx frac{2(n-1)}{2n-1} xx frac{2(n-2)}{2n-3} xx … xx frac{2}{3} xx 1`
  `= frac{2n}{2n+1} xx frac{2n}{2n} xx frac{2(n-1)}{2n-1} xx frac{2(n-1)}{2n-2} xx … xx frac{2}{3} xx frac{2}{2} xx 1`
  `= frac{2^n (n xx (n-1) xx .. xx 1) xx 2^n (n xx (n – 1) xx … xx 1)}{(2n + 1)!}`
  `= frac{2^(2n) (n!)^2}{(2n + 1)!}`
♦♦♦ Mean mark (iii) 16%.

 

iii.   `J_n = int_0^1 x^n (1-x)^n\ dx ,  \ n = 0, 1, 2, …`

`text{Let} \ \ x` `= sin^2 theta`
`frac{dx}{d theta}` `= 2 sin theta \ cos theta \ => \ dx = 2 sin theta \ cos theta \ d theta`

 

`text{When}`    `x = 0 \ ,` ` \ theta = 0`
  `x = 1  \ ,` ` \ theta = frac{pi}{2}`

 

`J_n` `= int_0^(frac{pi}{2}) (sin^2 theta)^n (1 – sin^2 theta)^n * 2 sin theta \ cos theta \ d theta`
  `= int_0^(frac{pi}{2}) sin^(2n) theta \ cos^(2n) theta * sin (2 theta)\ d theta`
  `= frac{1}{2^(2n)} int_0^(frac{pi}{2}) 2^(2n) sin^(2n) theta \ cos^(2n) theta * sin (2 theta)\ d theta`
  `= frac{1}{2^(2n)} int_0^(frac{pi}{2}) sin^(2n) (2 theta) * sin (2 theta)\ d theta`
  `= frac{1}{2^(2n)} int_0^(frac{pi}{2}) sin^(2n+1) (2 theta)\ d theta`
  `= frac{1}{2^(2n)} * frac{2^(2n) (n!)^2}{(2n+1)!}\ \ \ text{(using part (ii))}`
  `= frac{(n!)^2}{(2n + 1)!}`
♦♦♦ Mean mark (iv) 10%.

 

iv.   `text{If} \ \ I_n ≤ 1,`

`2^(2n) (n!)^2` ` ≤ (2n + 1)!`
`(2^n n!)^2` `≤ (2n + 1)!`

  
`text{Show} \ \ I_n ≤ 1 :`

`text{Consider the graphs}`

`y = sin(2 theta) \ \ text{and}\ \ y = sin^(2n + 1) (2 theta) \ \ text{for} \ \ 0 ≤ theta ≤ frac{pi}{2}`
 

`int_0^(frac{pi}{2}) sin(2 theta)` `= [ – frac{1}{2} cos (2 theta) ]_0^(frac{pi}{2})`
  `= – frac{1}{2} cos \ pi + frac{1}{2} cos \ 0`
  `= 1`

 
`y = sin(2 theta) \ => \ text{Range} \ [0, 1] \ \ text{for}\ \ theta ∈ [0, frac{pi}{2}]`

`sin^(2n+1) (2 theta)` `≤ sin (2 theta) \ \ text{for}\ \ theta ∈ [0, frac{pi}{2}]`
`sin^(2n+1) (2 theta)` `≤ 1`
`I_n` `≤ 1`
`therefore (2^n n!)^2` `≤ (2n + 1) !`

Calculus, EXT2 C1 2002 HSC 2b

For  `n = 0, 1, 2,`...

let  `I_n = int_0 ^{(pi)/(4)} tan^(n) theta  d theta`.

  1.  Show that  `I _1 = (1)/(2) ln2`.   (1 mark)

  2.  Show that, for  `n >= 2`,
     
         `I_n + I_(n - 2) = (1)/(n-1)`.   (3 marks)

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  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `text(Proof)\ text{(See Worked Solutions)}`
Show Worked Solution

i.         `I_n = int_0 ^{(pi)/(4)} tan^n theta \ d theta`

`I_1` `= int_0 ^{(pi)/(4)} tan theta \ d theta`
  `= [ -ln cos x ]_0 ^{(pi)/(4)}`
  `= [- ln cos\ (pi)/(4) + ln cos 0]`
  `= -ln 2^{-(1)/(2)}`
  `= (1)/(2) ln 2`

 

ii. `I_n` `= int_0 ^{(pi)/(4)} tan^(n – 2) theta ⋅ tan^2 theta \ d theta`
    `= int_0 ^{(pi)/(4)} tan^(n-2) theta ⋅ (sec^2 theta – 1) d theta`
    `= int_0 ^{(pi)/(4)} tan^(n – 2) theta ⋅ sec^2 theta \ d theta – int_0 ^{(pi)/(4)} tan^(n – 2) theta \ d theta`

 
`text(Let) \ \ u = tan theta`

`(du)/(d theta) = sec^2 theta \ => \ du = sec^2 theta \ d theta`

`text(When)` `theta` `= (pi)/(4),`   `\ \ \ \u` `= 1`
  `theta` `=0,`   `u` `= 0`

 

`I_n` `= int_0 ^1 u^(n-2) \ du – I_(n – 2)`
  `= [(1)/(n-1) ⋅ u^(n-1)]_0 ^1 – I_(n – 2)`
  `= (1)/(n – 1) – I_(n-2)`

 
`:. \ I_n + I_(n – 2)=(1)/(n – 1)\ \ \ \ (n>=2)`

Calculus, EXT2 C1 2008 HSC 3c

For  `n >= 0`, let  `I_n = int_0 ^{(pi)/(4)} tan^(2n) theta  d theta`.   

  1. Show that for `n >= 1`,
     
         `I _n = (1)/(2n - 1) - I_(n-1)`.  (2 marks)

  2. Hence, or otherwise, calculate  `I_3`.   (2 marks)

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  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `(52 – 15 pi)/(60)`
Show Worked Solution
i. `I_n` `= int_0 ^{(pi)/(4)} tan^(2n-2) theta * tan^2 theta \ d theta`
    `= int_0 ^{(pi)/(4)} tan^(2n-2) theta * (sec^2 theta – 1) \ d theta`
    `= int_0 ^{(pi)/(4)} tan^(2n-2) theta * sec^2 theta  d theta – int_0 ^{(pi)/(4)} tan^(2n-2) theta \ d theta`
    `= int_0 ^{(pi)/(4)} tan^(2n-2)theta * sec^2 theta \ d theta – I_(n-1)`

 

`text(Let) \ \ u = tan theta`

`(du)/(d theta) = sec^2 theta \ => \ du = \ sec^2 theta \ d theta`

`text(When)`    `theta` `= (pi)/(4)` `,` `\ \ \ \u` `= 1`
  `theta` `= 0` `,` `u` `= 0`

 

`I_n` `= int_0 ^1 u^(2n – 2) \ du – I_(n – 1)`
  `= [1/(2n -1) ⋅ u^(2n – 1)]_0 ^1 – I_(n – 1)`
  `= 1/(2n-1)(1^(2n-1)-0)  – I_(n – 1)`
  `= 1/(2n-1)  – I_(n – 1)`

 

ii.        `I_0 = int_0 ^{(pi)/(4)} d theta = (pi)/(4)`

`I_3` `= (1)/(5) – I_2`
  `= (1)/(5) – ((1)/(3) – I_1)`
  `= (1)/(5) – (1)/(3) + (1 – I_0)`
  `= (13)/(15) – (pi)/(4)`
  `= (52 – 15 pi)/(60)`

Calculus, EXT2 C1 2006 HSC 7b

  1. Let  `I_n = int_0^x sec^n t\ dt`,  where  `0 <= x <= pi/2`.  
     
    Show that   `I_n = (sec^(n - 2) x tan x)/(n - 1) + (n - 2)/(n - 1) I_(n - 2).`  (3 marks)

  2. Hence find the exact value of
     
         `int_0^(pi/3) sec^4 t\ dt.`  (2 marks)

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  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `2 sqrt 3`
Show Worked Solution
STRATEGY: The choice of `u=sec^(n-2)t` and `v=sec^2 t` was extremely important in being able to answer this question. Take note!
i.   `I_n` `=int_0^x sec^n t\ dt,\ \ \ 0 <= x < pi/2`
  `=int_0^x sec^(n – 2)t sec^2 t\ dt`

`text(Integrating by parts)`

`u` `=sec^(n-2)t,` `u′` `=(n-2) sec^(n-2)t tan\ t`
`v` `=tan\ t,` `v′` `=sec^2 t`
`­I_n` `=[tan t sec^(n – 2)t]_0^x – int_0^x tan t (n – 2) sec^(n – 3) t sec t tan t\ dt`
  `=tan x sec^(n – 2) x – (n – 2) int_0^x sec^(n – 2) t tan^2 t\ dt`
  `=tan x sec^(n – 2) x – (n – 2) int_0^x sec^(n – 2) t (1+sec^2 t)\ dt`
  `=tan x sec^(n – 2) x – (n – 2) int_0^x sec^n t\ dt + (n – 2) int_0^x sec^(n – 2)t\ dt`
  `=tan x sec^(n – 2) x-(n-2)I_n+(n-2)I_(n-2)`

 

`I_n + (n – 2) I_n` `= tan x sec^(n – 2) x + (n – 2) I_(n – 2)`
`(n – 1)I_n` `= tan x sec^(n – 2) x + (n – 2) I_(n – 2)`
`:.I_n ` `= (tan x sec^(n – 2) x)/(n – 1) + ((n – 2)/(n – 1)) I_(n – 2)`

 

ii.   `int_0^(pi/3) sec^4 t\ dt ­=` `(tan\ pi/3 sec^2\ pi/3)/3 + 2/3 int_0^(pi/3) sec^2 t\ dt`
`­=` `(sqrt 3 xx 4)/3 + 2/3 [tan t]_0^(pi/3)`
`­=` `(4 sqrt 3)/3 + 2/3 (sqrt 3 – 0)`
`­=` `2 sqrt 3`