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Calculus, EXT2 C1 2020 HSC 13d*

  1. By expanding `(text{cis}\theta + text{cis}(-theta))^4` show that
  2.    `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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a.    `text{See Worked Solution}`

b.    `frac{3 pi}{16}`

Show Worked Solution

a.    `text{cis}\theta + text{cis}(-theta) = 2 cos theta\ \ …\ (1)`

`(text{cis}\theta + text{cis}(-theta))^4= 16 cos^4(4theta)`

`text{Expand LHS:}`

`(text{cis}\theta + text{cis}(-theta))^4`

`= text{cis}(4theta)+4text{cis}(2theta)+6+4text{cis}(-2theta)+text{cis}(4theta)`

`= 2text{cos}(4theta)+8text{cos}(2theta)+6\ \ text{(using (1) above)}`
 

`text{Equating sides:}`

`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)`

 

b.     `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 2025 HSC 11f

Find \(\displaystyle \int \dfrac{5}{\sqrt{7-x^2-6 x}} \, dx\).   (2 marks)

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\(5\,\sin^{-1} \left( \dfrac{x+3}{4} \right) +c\)

Show Worked Solution
\(\displaystyle \int \dfrac{5}{\sqrt{7-x^2-6 x}} \, dx\) \(=\displaystyle \int \dfrac{5}{\sqrt{16-(x^2+6x+9)}} \, dx\)  
  \(=\displaystyle \int \dfrac{5}{\sqrt{16-(x+3)^2}} \, dx\)  
  \(=5\,\sin^{-1} \left( \dfrac{x+3}{4} \right) +c\)  

Calculus, EXT2 C1 2024 HSC 11d

Evaluate  \(\displaystyle\int_0^{\frac{\pi}{2}} \dfrac{1}{\sin \theta+1}\, d \theta\).   (3 marks)

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\(1\)

Show Worked Solution

\(t=\tan (\frac{\theta}{2}), \ \sin \theta=\dfrac{2 t}{1+t^2}\)

\(d t=\dfrac{1}{2} \sec ^2 (\frac{\theta}{2})\, d \theta \ \Rightarrow \ d \theta=\dfrac{2}{1+\tan ^2 (\frac{\theta}{2})}\, d t=\dfrac{2}{1+t^2}\, d t\)

\(\text {When}\ \ \theta=\dfrac{\pi}{2}\ \ \Rightarrow\ \ \tan (\frac{\theta}{2})=1\)

\(\text {When}\ \ \theta=0\ \ \Rightarrow\ \ \tan(\frac{\theta}{2})=0\)

\(\displaystyle{\int}_0^{\frac{\pi}{2}} \dfrac{1}{\sin \theta+1}\, d \theta\) \(=\displaystyle{\int}_0^1 \frac{1}{\dfrac{2 t}{1+t^2}+1} \cdot \frac{2}{1+t^2}\, d t\)
  \(=\displaystyle{\int}_0^1 \dfrac{2}{2 t+1+t^2}\, d t\)
  \(=\displaystyle{\int}_0^1 \dfrac{2}{(t+1)^2}\, d t\)
  \(=-\left[\dfrac{2}{t+1}\right]_0^1\)
  \(=-\left(\dfrac{2}{2}-2\right)\)
  \(=1\)

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

Show Worked Solution

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

 
\(\text{Let}\ \ u=5-4x-x^2 \)

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

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

Calculus, EXT2 C1 2022 HSC 11f

Using the substitution  `t=tan\ x/2`, find

`int(dx)/(1+cos x-sin x)`   (3 marks)

 

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`-lnabs(1-tan(x/2))+C`

Show Worked Solution

`text(Let)\ \ t = tan\ x/2, \ cos\ x = (1-t^2)/(1 + t^2), \ sin\ x=(2t)/(1+t^2)`

`dt = 1/2 sec^2\ x/2\ dx \ => \ d x = (2\ dt)/(sec^2\ x/2) = 2/(1 + t^2)\ dt`

`text{I}` `= int(dx)/(1+cos x-sin x)`  
  `=int 1/(1+(1-t^2)/(1 + t^2)-(2t)/(1 + t^2)) *2/(1 + t^2)\ dt`  
  `=int 2/(1+t^2+1-t^2-2t)\ dt`  
  `=int 1/(1-t)\ dt`  
  `=-ln abs(1-t)+C`  
  `=-lnabs(1-tan(x/2))+C`  

Calculus, EXT2 C1 2022 HSC 4 MC

Of the following expressions, which one need NOT contain a term involving a logarithm in its anti-derivative?

  1. `(x+2)/(x^(2)+4x+5)`
  2. `(x+2)/(x^(2)-4x-5)`
  3. `(x-1)/(x^(3)-x^(2)+x-1)`
  4. `(x+1)/(x^(3)-x^(2)+x-1)`
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`C`

Show Worked Solution

`text{Consider the denominator of}\ C:`

`x^(3)-x^(2)+x-1` `=x^2(x-1)+(x-1)`  
  `=(x^2+1)(x-1)`  

 
`(x-1)/(x^(3)-x^(2)+x-1)=(x-1)/((x^2+1)(x-1))=1/(x^2+1)`

`int 1/(x^2+1)\ dx=tan^(-1)(x)+c`

`=>C`

Calculus, EXT2 C1 2021 SPEC2 2

Evaluate  `int_0^1 (2x + 1)/(x^2 + 1)\ dx`.  (3 marks)

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`text(See Worked Solutions)`

Show Worked Solution
`int_0^1 (2x + 1)/(x^2 + 1)\ dx` `= int_0^1 (2x)/(x^2 + 1)\ dx + int_0^1 1/(x^2 + 1)\ dx`
  `= [log_e(x^2 + 1)]_0^1 + [tan^(-1)(x)]_0^1`
  `= log_e 2 – log_e 1 + tan^(-1)(1) – tan^(-1)(0)`
  `= log_e 2 + pi/4`

Calculus, EXT2 C1 2021 HSC 14a

Evaluate  `int_0^(pi/2) 1/(3 + 5cosx)\ dx`.  (4 marks)

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`1/4 ln 3`

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`text(Let)\ \ t = tan\ theta/2, \ cos theta = (1-t^2)/(1 + t^2)`

`dt = 1/2 sec^2\ theta/2\ d theta \ => \ d theta = (2\ dt)/(sec^2\ theta/2) = 2/(1 + t^2)\ dt`

`text(Convert limits:)`

`theta` `= pi/2  → \ t=1`
`theta` `= 0  → \ t=0`
`int_0^(pi/2) 1/(3 + 5cosx)\ dx` `= int_0^1 1/(3 + 5((1-t^2)/(1 + t^2))) · 2/(1 + t^2)\ dt`
  `= int_0^1 2/(3(1 + t^2) + 5(1-t^2))\ dt`
  `= int_0^1 1/(4-t^2)\ dt`
  `= int_0^1 1/((2 + t)(2-t))\ dt`
  `= 1/4 int_0^1 1/(2+t) + 1/(2-t)\ dt`
  `= 1/4 [ln |2 + t|-ln|2-t|]_0^1`
  `= 1/4 [ln 3-ln1-(ln 2-ln 2)]`
  `= 1/4 ln 3`

Calculus, EXT2 C1 2005 HSC 1e

Let  `t=tan(theta/2).`

  1. Show that  `(dt)/(d theta) = 1/2(1+t^2)`   (1 mark)

  2. Show that  `sin theta = (2t)/(1+t^2).`   (2 marks)

  3. Use the substitution  `t=tan(theta/2)`  to find  `int text(cosec)\ theta\ d theta.`   (2 marks)

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a.    `text(See Worked Solutions)`

b.    `text(See Worked Solutions)`

c.    `log_e | tan frac{theta}{2} | + c`

Show Worked Solution
a.     `t` `= tan frac{theta}{2}`
  `frac{dt}{d theta}` `= frac{1}{2} text{sec}^2 frac{theta}{2}`
    `= frac{1}{2} (1 + tan^2 frac{theta}{2})`
    `= frac{1}{2} (1 + t^2)`

 

b.    `text{Show} \ \ sin theta = frac{2t}{1 + t^2} :`
 

`sin theta` `= 2 \ sin frac{theta}{2} cos frac{theta}{2}`
  `= 2 * frac{t}{sqrt(1 + t^2)} * frac{1}{sqrt(1 + t^2)}`
  `= frac{2t}{1 + t^2}`

 

c.    `int \ text{cosec} \ theta \ d theta`

`t = tan frac {theta}{2}`

`frac{dt}{d theta} = frac{1}{2} text{sec}^2 frac{theta}{2} \ , \ d theta = frac{2dt}{sec^2 frac{theta}{2}} = frac{2}{1 + t^2}  dt`

`int \ text{cosec} \ theta\ d theta` `= int frac{1 + t^2}{2t} xx frac{2}{1 + t^2} dt`
  `= int frac{1}{t}\ dt`
  `= log_e | t | + c`
  `= log_e | tan frac{theta}{2} | + c`

Calculus, EXT2 C1 2020 HSC 6 MC

Which expression is equal to `int frac{1}{x^2 + 4x + 10}\ dx`?

  1. `frac{1}{sqrt(6)} tan^-1 (frac{x + 2}{sqrt{6} )) + c`
  2. `tan^-1 (frac{x + 2}{sqrt{6} )) + c`
  3. `frac{1}{2 sqrt(6)} ln | frac{x + 2 - sqrt(6)}{x + 2 + sqrt(6)} | + c`
  4. `ln | frac{x + 2 - sqrt(6)}{x + 2 + sqrt(6)} | + c`
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`A`

Show Worked Solution
`int frac{1}{x^3 + 4x + 10}\ dx` `= int frac{1}{(x + 2)^2 + (sqrt6)^2}\ dx`
  `= frac{1}{6} tan^-1 (frac{x + 2}{sqrt6}) + c`

`=> \ A`

Calculus, EXT2 C1 EQ-Bank 12

By completing the square and using the table of standard integrals, find

`int(dx)/(4x^2-4x+10)`   (2 marks)

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`1/6 tan^-1((2x-1)/(3))+C`

Show Worked Solution
`int(dx)/(4x^2-4x+10)` `=int(dx)/(3^2+(2x-1)^2)`
  `=1/2 int 2/(3^2+(2x-1)^2) \ dx`
  `=1/6 tan^-1((2x-1)/(3))+C`

Calculus, EXT2 C1 2019 HSC 7 MC

Which of these integrals has the largest value?

  1. `int_0^(pi/4) tan x\ dx`
  2. `int_0^(pi/4) tan^2 x\ dx`
  3. `int_0^(pi/4) 1 - tan x\ dx`
  4. `int_0^(pi/4) 1 - tan^2 x\ dx`
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`D`

Show Worked Solution

`text(Consider options A and B:)`

`text(Consider options C and D:)`


 

`:. int_0^(pi/4) 1 – tan^2 x\ dx\ \ text(is the largest)`

`text{(largest area under the curve}`

  `text(between)\ \ x=0 \ and \ x=pi/4. text{)}`

 
`=>   D`

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`

Show Worked Solution

`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 2018 HSC 14a

Using the substitution  `t = tan\ theta/2`  evaluate  `int_0^(pi/2) (d theta)/(2 - costheta)`.  (3 marks)

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`(2sqrt3 pi)/9`

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`t = tan\ theta/2, \ costheta = (1 – t^2)/(1 + t^2), \ d theta = 2/(1 + t^2) dt`

`text(When)\ \ theta = pi/2, \ t = 1`

`text(When)\ \ theta = 0, \ t = 0`
 

`int_0^(pi/2) (d theta)/(2 – costheta)` `= int_0^1 1/(2 – (1 – t^2)/(1 + t^2)) · 2/(1 + t^2)\ dt`
  `= int_0^1 2/(2(1 + t^2) – (1 – t^2))\ dt`
  `= int_0^1 2/(1 + 3t^2)\ dt`
  `= [2/sqrt3 tan^(−1)(sqrt3 t)]_0^1`
  `= 2/sqrt3 tan^(−1) sqrt3 – 0`
  `= (2pi)/(3sqrt3)`
  `= (2sqrt3 pi)/9`

Calculus, EXT2 C1 2018 HSC 12c

Find  `int(x^2 + 2x)/(x^2 + 2x + 5)\ dx`.  (3 marks)

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`x – 5/2 tan^(−1) ((x + 1)/2) + c`

Show Worked Solution
`int(x^2 + 2x)/(x^2 + 2x + 5)\ dx` `= int((x^2 + 2x + 5) – 5)/(x^2 + 2x + 5)\ dx`
  `= int 1 – 5/(x^2 + 2x + 5)\ dx`
  `= int 1 – 5/(2^2 + (x + 1)^2)\ dx`
  `= x – 5/2 tan^(−1) ((x + 1)/2) + c`

Calculus, EXT2 C1 2017 HSC 11d

Using the substitution  `t = tan {:theta/2:}`, or otherwise, evaluate

`int_0^((2 pi)/3) 1/(1 + cos theta)\ d theta`.  (3 marks)

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`sqrt 3`

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`text(Let)\ \ t = tan {:theta/2:}, \ cos theta = (1 – t^2)/(1 + t^2), \ d theta = (2 dt)/(1 + t^2)`

 

`text(When)\ theta = 0, t = tan 0 = 0`

`text(When)\ theta = (2 pi)/3, t = tan {:pi/3:} = sqrt 3`

`int_0^((2pi)/3) 1/(1 + cos theta)\ d theta` `= int_0^(sqrt 3) ((2dt)/(1 + t^2))/((1 + t^2)/(1 + t^2) + (1 – t^2)/(1 + t^2))`
  `= int_0^(sqrt 3) (2/(1 + t^2))/(2/(1 + t^2))\ dt`
  `= int_0^(sqrt 3) 1\ dt`
  `= [t]_0^ (sqrt 3)`
  `= sqrt 3 – 0`
  `= sqrt 3`

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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a.    `text(See Worked Solutions)`

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

c.    `(3 pi)/16`

Show Worked Solution

a.   `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`

 

b.    `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)`

 

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

b.     `2 ln\ |sin\ theta/2| + c`

Show Worked Solution
a.   `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%.
b.   `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 2006 HSC 1e

Use the substitution  `t = tan\ theta/2`  to show that

`int_(pi/2)^((2 pi)/3) (d theta)/(sin theta) = 1/2 log 3.`  (3 marks)

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`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

`text(Let)\ \ t = tan\ \ theta/2,\ \ sin theta = (2t)/(1 + t^2),\ \ d theta = 2/(1 + t^2)\ dt`

`text(When)\ \ theta = pi/2,\ \ t = 1`

`text(When)\ \ theta = (2 pi)/3,\ \ t = sqrt 3`

`:.int_(pi/2)^((2 pi)/3) (d theta)/(sin theta)` `= int_1^sqrt 3 (1 + t^2)/(2t) xx2/(1 + t^2)\ dt`
  `= int_1^ sqrt 3 (dt)/t`
  `= [ln t]_1^sqrt 3`
  `= ln sqrt 3 – ln 1`
  `=ln3^(1/2)-0`
  `= 1/2 ln 3`

Calculus, EXT2 C1 2009 HSC 1c

Find  `int x^2/(1 + 4x^2)\ dx.`  (3 marks)

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`x/4 – 1/8 tan^-1 2x + c`

Show Worked Solution
`x^2/(1 + 4x^2)` `= 1/4 xx (4x^2)/(1 + 4x^2)`
  `= 1/4 xx (1 + 4x^2)/(1 + 4x^2) – 1/4 xx 1/(1 + 4x^2)`
  `=1/4-1/4 xx 1/(1 + 4x^2)`

 

`:.int x^2/(1 + 4x^2)\ dx` `= 1/4 int 1\ dx – 1/4 int 1/(1 + 4x^2)\ dx`
  `= x/4 – 1/4 xx 1/2 tan^-1 2x + c`
  `= x/4 – 1/8 tan^-1 2x + c`

Calculus, EXT2 C1 2010 HSC 1d

Using the substitution  `t = tan\ x/2`, or otherwise, evaluate  `int_0^(pi/2) (dx)/(1 + sin\ x)`.   (4 marks)

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`1`

Show Worked Solution

`t = tan\ x/2, \ dx = (2\ dt)/(1 + t^2), \ sin\ x = (2t)/(1 + t^2)`

`text(When)\ \ x=pi/2,\ \ t=tan\ pi/4=1`

`text(When)\ \ x=0,\ \ t=tan0=0`

`:.int_0^(pi/2) (dx)/(1 + sin\ x)` `=int_0^1 ((2\ dt)/(1 + t^2))/(1 + (2t)/(1 + t^2))`
  `=int_0^1 2/(1 + t^2 + 2t)\ dt`
  `=int_0^1 (2)/((1 + t)^2)\ dt`
  `=[(-2)/(1 + t)]_0^1`
  `=(-2)/2 − (-2)/1`
  `=1`

Calculus, EXT2 C1 2010 HSC 1b

Evaluate  `int_0^(pi/4) tan\ x\ dx`.   (3 marks) 

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`1/2 ln 2 \ \ text(or)\ \ ln\ sqrt2`

Show Worked Solution
`int_0^(pi/4) tan\ x\ dx` `=int_0^(pi/4) (sin\ x)/(cos\ x)\ dx`
  `=[-ln\ cos\ x]_0^(pi/4)`
  `=[-ln\ cos\ pi/4 – (-ln cos 0)]`
  `=-ln\ 1/sqrt2 + ln\ 1`
  `=ln sqrt2`
  `=1/2 ln 2`

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

b.     `1/4 log_e 3`

Show Worked Solution

a.    `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`

 

b.    `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 1e

Evaluate  `int_-1^1 1/(5 - 2t + t^2) \ dt.`  (3 marks)

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`pi/8`

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`int_-1^1 1/{(5 – 2t + t^2)}dt` `= int_-1^1 1/{(4 + 1 – 2t + t^2)dt}`
  `= int_-1^1 1/(4 + (t – 1)^2)dt`
  `= 1/2[tan^-1 ((t – 1)/2)]_-1^1`
  `= 1/2 [tan^-1 0 – tan^(-1)(-1)]`
  `= 1/2 [0 – (-pi/4)]`
  `= pi/8`

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`

Calculus, EXT2 C1 2012 HSC 12a

Using the substitution  `t = tan\ theta/2`, or otherwise, find  `int(d theta)/(1 − cos\ theta)`.   (3 marks)

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`-cot\ theta/2 + c`

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`t = tan\ theta/2`

`dt=1/2 sec^2 (theta/2)\ d theta,\ \ \ d theta=(2\ dt)/sec^2 (theta/2)=2/(1+t^2)\ dt`

`cos\ theta = (1 − t^2)/(1 + t^2)`

`int(d theta)/(1 − cos\ theta)` `= int1/(1 − ((1 − t^2)/(1 + t^2))) xx 2/(1 + t^2)\ dt`
  `= int 2/(1 + t^2 − (1 − t^2))`
  `= int(dt)/(t^2)`
  `= -1/t + c`
  `= -1/(tan\ theta/2) + c`
  `= -cot\ theta/2 + c`

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

b.    `tan theta + tan^3 theta + 3/5 tan^5 theta + 1/7 tan^7 theta + C`

Show Worked Solution
a.     `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%.

b.    `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` 

Calculus, EXT2 C1 2013 HSC 12a

Using the substitution  ` t = tan\ x/2`, or otherwise, evaluate

`int_0^(pi/2) 1/(4 + 5 cos x)\ dx.`  (4 marks)

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`1/3 ln 2`

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`t = tan\ x/2`

`=>cos x = (1 – t^2)/(1 + t^2)\ ,\ \ \ dx = (2\ dt)/(1 + t^2)`

`text(When)\ \ x = 0\ ,\ t = 0\ ;\ \ x = pi/2\ ,\ t = 1`

`int_0^(pi/2) (dx)/(4 + 5 cos x)` `= int_0^1 1/{4 + (5(1 – t^2))/(1 + t^2)} xx (2\ dt)/(1 + t^2)`
  `= int_0^1 (2\ dt)/(4 + 4t^2 + 5 – 5t^2)`
  `= int_0^1 (2\ dt)/(9 – t^2)`
  `=2 int_0^1 1/((3-t)(3+t))`
  `= 1/3 int_0^1 (1/(3 – t) + 1/(3 + t))\ dt`
  `= 1/3 [-ln (3 – t) + ln (3 + t)]_0^1`
  `= 1/3 [(-ln 2 + ln 4) – (-ln 3 + ln 3)]`
  `= 1/3 ln2`

Calculus, EXT2 C1 2014 HSC 13a

Using the substitution  `t = tan\ x/2`, or otherwise, evaluate

`int_(pi/3)^(pi/2) 1/(3sinx - 4cosx + 5)\ dx`.  (3 marks)

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`(2sqrt3 − 3)/6`

Show Worked Solution

`t = tan\ x/2,\ \ \ sinx=(2t)/(1+t^2)`

`cos x = (1-t^2)/(1+t^2),\ \ \ dx=(2 dt)/(1+t^2`

`text(When)\ \ \ x = pi/3,\ \ t = tan\ pi/6 =1/sqrt3`

`text(When)\ \ \ x = pi/2, \ \ t = tan\ pi/4=1`.

`int_(pi/3)^(pi/2) 1/(3\ sin\ x − 4\ cos\ x + 5)`
`= int_(1/sqrt3)^1 1/((6t)/(1 + t^2) − (4(1 − t^2))/(1 + t^2) + 5) xx (2dt)/(1 + t^2)`
`= int_(1/sqrt3)^1 2/(6t − 4 + 4t^2 + 5 + 5t^2)dt`
`= int_(1/sqrt3)^1 2/(9t^2 + 6t + 1)dt`
`= 2 int_(1/sqrt3)^1 (dt)/((3t + 1)^2)`
`= -2/3[1/(3t + 1)]_(1/sqrt3)^1`
`= -2/3(1/4 − 1/(sqrt3 + 1))` 
`= -2/3(1/4 − (sqrt3 − 1)/2)`
`= -1/6+sqrt3/3-1/3`
`= (2sqrt3 − 3)/6`