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Calculus, EXT2 C1 2025 HSC 12a

Using integration by parts, evaluate \(\displaystyle \int_0^{\small{\dfrac{\pi}{2}}} x\, \sin x \, dx\).   (3 marks)

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\(\displaystyle \int_0^{\small{\dfrac{\pi}{2}}} x\, \sin x \, dx = 1\)

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\(u\) \(=x\) \(u^{\prime}\) \(=1\)
\(v^{\prime}\) \(=\sin\,x\) \(v\) \(=-\cos\,x\)
\(\displaystyle \int_0^{\small{\dfrac{\pi}{2}}} x\, \sin x \, dx\) \(=\Big[-x\,\cos\,x\Big]_0^{\small{\dfrac{\pi}{2}}} + \displaystyle \int_0^{\small{\dfrac{\pi}{2}}} \cos x \, dx\)  
  \(=(0-0)+\Big[\sin\,x\Big]_0^{\small{\dfrac{\pi}{2}}} \)  
  \(=\sin\,\dfrac{\pi}{2}-\sin\,0\)  
  \(=1\)  

Calculus, EXT2 C1 2024 HSC 11a

Find \(\displaystyle \int x e^x\, d x\)   (2 marks)

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\(x e^x-e^x+c\)

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\(u=x \quad \ \ u^{\prime}=1\)

\(v^{\prime}=e^x \quad v=e^x\)

  \(\displaystyle\int x e^x \,d x\) \(=u v^{\prime}-\displaystyle \int v u^{\prime}\, d x\)
    \(=x e^x- \displaystyle \int e^x \cdot 1\, d x\)
    \(=x e^x-e^x+c\)

Calculus, EXT2 C1 2022 HSC 15c

Using the substitution  `x=tan^(2)theta`, evaluate

          `int_(0)^(1)sin^(-1)sqrt((x)/(1+x))\ dx`   (4 marks)

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`pi/2-1`

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`x=tan^(2)theta`

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

`text{When}\ x=0, \ theta=0`

`text{When}\ x=1, \ theta=pi/4`


Mean mark 55%.
`sin^(-1)sqrt((x)/(1+x))` `=sin^(-1)sqrt((tan^(2)theta)/(1+tan^(2)theta))`  
  `=sin^(-1)sqrt((tan^(2)theta)/(text{sec}^(2)theta))`  
  `=sin^(-1)sqrt((sin^(2)theta))`  
  `=sin^(-1)(sintheta)`  
  `=theta`  

 

`int_(0)^(1)sin^(-1)sqrt((x)/(1+x))\ dx=int_(0)^(pi/4)\ theta xx 2sec^2theta\ tantheta\ d theta`
 

`text{Integrating by parts:}`

`u` `=theta` `u^{′}` `=1`
`v^{′}` `=2tan theta\ sec^2 theta` `v` `=tan^(2)theta`

 
`int_(0)^(pi/4)\ theta xx 2sec^2theta\ tantheta\ d theta`

`=[theta\ tan^(2)theta]_0^(pi/4)-int_0^(pi/4)tan^(2)theta\ d theta`

`=(pi/4 xx 1 -0)-int sec^(2)theta-1\ d theta`

`=pi/4-[tan theta-theta]_0^(pi/4)`

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

`=pi/2-1`

Calculus, EXT2 C1 2021 HSC 2 MC

Which expression is equal to  `int x^5 e^{7x} dx`?

  1. `1/7 x^5 e^{7x} - 5/7 int x^4 e^{7x} dx`
  2. `1/7  x^5 e^{7x} - 5/7 int x^5 e^{7x} dx`
  3. `5/7 x^4 e^{7x} - 5/7 int x^4 e^{7x} dx`
  4. `5/7  x^4 e^{7x} - 5/7 int x^5 e^{7x} dx`
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`A`

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`u = x^5`   `v^{′} = e^{7x}`
`u^{′} = 5x^4`   `v = 1/7 e^{7x}`
`int uv^{′}\ dx` `= uv-int u^{′}v \ dx`  
  `= 1/7  x^5 e^{7x}-5/7 int x^4 e^{7x}\ dx`  

 
`=>\ A`

Calculus, EXT2 C1 2020 HSC 11b

Use integration by parts to evaluate  `int_1^e x ln x \ dx`.   (3 marks)

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`frac{e^2 + 1}{4}`

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`u = ln \ x` `v′ = x`
`u′ = frac{1}{x}` `v = frac{x^2}{2}`

 

`int _1^e x \ ln \ x \ dx` `= [ frac{x^2}{2} · ln \ x ]_1^e – int_1^e frac{x^2}{2} · frac{1}{x}\ dx`
  `= [frac{e^2}{2} ln \ e – frac{1}{2} ln 1]- int_1^e frac{x}{2}\ dx`
  `= frac{e^2}{2} – [ frac{x^2}{4}]_1^e`
  `= frac{e^2}{2} – ( frac{e^2}{4} – frac{1}{4} )`
  `= frac{e^2 + 1}{4}`

Calculus, EXT2 C1 2005 HSC 1c

Use integration by parts to evaluate  `int_1^e x^7 log_e x  dx`.   (3 marks)

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`(7e^8 – 1)/(64)`

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  `u` `= log_e x,` `\ \ \ \ u^(′)` `= (1)/(x)`
  `v^(′)` `= x^7,` `v` `= (1)/(8) x^8`

 

`int_1^e x^7\ log_e x  dx` `= uv-int u^(′) v \ dx`
  `= [(x^8)/(8) ⋅ log_e x]_1 ^e-(1)/(8) int_1 ^e (1)/(x) ⋅ x^8 dx`
  `= ((e^8)/(8) log_e e-(1)/(8) log_e 1)-(1)/(8)[(1)/(8) x^8]_1 ^e`
  `= (e^8)/(8)-(1)/(8) ((e^8)/(8)-(1)/(8))`
  `= (e^8)/(8)-((e^8-1)/(64))`
  `= (7e^8 +1)/(64)`

Calculus, EXT2 C1 2003 HSC 1b

Use integration by parts to find   `int x^3 log_e x  dx`   (3 marks)

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`(x^4 log_e x)/(4)-(x^4)/(16) + C`

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`u` `= log_e x,` `\ \ \ \ u^{′}` `= (1)/(x)`
`v^{′}` `= x^3,` `v` `= (x^4)/(4)`

 

`int x^3 log_e x  dx` `= uv-int u^{′}v \ dx`
  `= (x^4)/(4)  log_e x-int (1)/(x) ⋅ (x^4)/(4)  dx`
  `= (x^4 log_e x)/(4)-(1)/(4) int x^3 dx`
  `= (x^4 log_e x)/(4)-(x^4)/(16) + C`

Calculus, EXT2 C1 2004 HSC 1a

Use integration by parts to find  `int x e^(3x)  dx`.   (2 marks)

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`(x e^(3x))/3-(1)/(9)  e^(3x) + C`

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`u` `= x` `\ \ \ \ u^{′}` `= 1`
`v^{′}` `= e^(3x)` `v` `= (1)/(3)  e^(3x)`

 

`int x e^(3x)  dx` `= uv-int u^{′} v \ dx`
  `= (xe^(3x))/(3)-(1)/(3) int e ^(3x)  dx`
  `= (x e^(3x))/3-(1)/(9)  e^(3x) + C`

Calculus, EXT2 C1 2019 HSC 3 MC

Which expression is equal to  `int x cos x\ dx`?

  1. `-x sin x + cos x + C`
  2. `-x sin x - cos x + C`
  3. `x sin x + cos x + C`
  4. `x sin x - cos x + C`
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`C`

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`u = x` `v prime = cos x`
`u prime = 1` `v = sin x`

 

`int uv prime\ dx` `= uv – int u prime v\ dx`
  `= x sin x – int sin x\ dx`
  `= x sin x + cos x + C`

 
`=>   C`

Calculus, EXT2 C1 2017 HSC 12c

Find  `int x tan^(-1) x\ dx`.  (3 marks)

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`1/2(x^2 tan^(-1)x – x + tan^(-1) x) + c`

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`I = int x tan^(-1) x\ dx`

`text(Let)\ \ \ u` `= tan^(-1) x`   `v prime` `= x`
`u prime` `= 1/(1 + x^2)`   `v` `= x^2/2`

 

`I` `= uv – int u prime v\ dx`
  `= tan^(-1) x · x^2/2 – int 1/(1 + x^2) · x^2/2\ dx`
  `= x^2/2  tan^(-1) x – 1/2 int x^2/(1 + x^2)\ dx`
  `= x^2/2  tan^(-1) x – 1/2 int (1 + x^2 – 1)/(1 + x^2)\ dx`
  `= x^2/2  tan^(-1) x – 1/2 int 1 – 1/(1 + x^2)\ dx`
  `= x^2/2  tan^(-1) x – 1/2 [x – tan^(-1) x] + c`
  `= x^2/2  tan^(-1) x – 1/2 x + 1/2 tan^(-1) x + c`
  `= 1/2(x^2 tan^(-1) x – x + tan^(-1) x) + c`

Calculus, EXT2 C1 2016 HSC 12b

  1. Differentiate  `x\ f(x)-int x\ f^(′)(x)\ dx.`  (1 mark)

  2. Hence, or otherwise, find  `int tan^-1 x\ dx.`  (2 marks)

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  1. `f(x)`
  2. `x tan^(−1)x-1/2 ln(1 + x^2)`
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i.   `d/dx (x\ f(x)-int x\ f^(′)(x)\ dx)`

`= x\ f^(′)(x) + f(x)-x\ fprime(x)`

`= f(x)`

 

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

Calculus, EXT2 C1 2016 HSC 11b

Find  `int x e^(-2x)\ dx.`  (3 marks)

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`-1/2 xe^(-2x)-1/4 e^(-2x) + c`

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`text(Integrating by parts:)`

`text(Let)` `u` `= x` `v^{′}` `= e^(-2x)`
  `u^{′}` `= 1` `v` `= -1/2e^(-2x)`

 

`int xe^(-2x)\ dx` `= x · -1/2 e^(-2x) + 1/2int e^(-2x)\ dx`
  `= -1/2 xe^(-2x)-1/4 e^(-2x) + c`

Calculus, EXT2 C1 2015 HSC 6 MC

Which expression is equal to  `int x^2 sin x\ dx`

  1. `-x^2 cos x - int 2 x cos x\ dx`
  2. `-2x cos x + int x^2 cos x\ dx`
  3. `-x^2 cos x + int 2x cos x\ dx`
  4. `-2x cos x - int x^2 cos x\ dx`
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`C`

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`u` `= x^2,` `\ \ \ \ u′` `= 2x`
`v′` `= sin x,` `v` `= -cos x`
`int uv′\ dx` `=uv-int u′v\ dx`
  `= x^2 (-cos x) – int 2x (-cos x) dx`
  `= -x^2 cos x + int 2x cos x\ dx`

`=>  C`

Calculus, EXT2 C1 2006 HSC 1d

Evaluate  `int_0^2 te^-t\ dt.`  (3 marks)

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`1-3/e^2`

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`u` `=t` `u^{′}` `=1`
`v` `=-e^-t\ dt` `v^{′}` `=e^-t`

 

`int_0^2 te^-t\ dt =` `[t (-e^-t)]_0^2-int_0^2 1 xx (-e^-t)\ dt`
 `­=` `[(-2e^-2)-0]-[e^-t]_0^2`
 `­=` `-2/e^2-(1/e^2 – 1)`
 `­=` `1-3/e^2`

Calculus, EXT2 C1 2014 HSC 16c

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

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`x/(1+ln x) +c`

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`I=int(ln\ x)/((1 + ln\ x)^2)\ dx`

♦♦ Mean mark 31%.
STRATEGY: This is as challenging as integration gets … a substitution followed by applying integration by parts. Note that simply substituting `u=ln x` gets you a full mark.
`text(Let)\ \ \ u=` `ln x`
`(du)/(dx)=` `1/x=1/e^u`
`du=` `1/e^u\ dx`
`dx=` `e^u\ du`
`I` `=int (u e^u)/(1+u)^2\ du`
  `=int ((1+u) e^u)/(1+u)^2\ du- int e^u/(1+u)^2\ du`
  `=int e^u/(1+u)\ du -int e^u/(1+u)^2\ du`
 

`text{Using integration by parts:}`

`u` `=-e^u`    `u′` `=-e^u`
`v′` `=-(1+u)^-2` `v` `=(1+u)^-1`
 `:.I` `=int e^u/(1+u)\ du – ((-e^u)/(1+u) + int (e^u)/(1+u)\ du)`
  `=e^u/(1+u) +c`
  `=x/(1+ln x) +c`

Calculus, EXT2 C1 2014 HSC 11b

Evaluate  `int_0^(1/2)(3x-1)\ cos\ (pix)\ dx`.  (3 marks)

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`(pi-6)/(2pi²)`

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`text(Integrating by parts:)`

`u` `=3x-1` `u^{′}` `=3`
`v^{′}` `=cos(pi x)` `v` `=1/pi sin(pi x)`

`int_0^(1/2)(3x-1)\ cos\ (pix)\ dx`

`= [(3x-1)(sin\ (pix))/pi]_0^(1/2)-int_0^(1/2) 3 xx (sin\ (pix))/pi\ dx`

`= (1/(2pi)\ sin\ pi/2-0) − 3/pi[(-cos\ (pix))/pi]_0^(1/2)`

`= 1/(2pi) + 3/(pi^2)(cos\ pi/2-cos\ 0)`

`= 1/(2pi) + 3/(pi^2)(0-1)`

`= 1/(2pi)-3/(pi^2)`