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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 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 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`

Show Worked Solution
`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 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)`