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Calculus, EXT1 C2 2024 HSC 13d

Using the substitution  \(u=e^x+2 e^{-x}\),  and considering \(u^2\), find  \(\displaystyle \int \frac{e^{3 x}-2 e^x}{4+8 e^{2 x}+e^{4 x}}\, d x\).   (3 marks)

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\(\displaystyle \frac{1}{2} \tan ^{-1}\left(\frac{e^x+2 e^{-x}}{2}\right)+c\)

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\(u=e^x+2 e^{-x} \ \Rightarrow \ u^2=\left(e^x+2 e^{-x}\right)^2=e^{2 x}+4+4 e^{-2 x}\)

\(\dfrac{du}{dx}=e^x-2 e^{-x} \ \Rightarrow \ du=\left(e^x-2 e^{-x}\right)\, d x\)

  \(\displaystyle \int \frac{e^{3 x}-2 e^x}{4+8 e^{2 x}+e^{4 x}}\, d x\)
    \(=\displaystyle \int \frac{e^{3 x}-2 e^x}{4+8 e^{2 x}+e^{4 x}} \times \frac{e^{-2 x}}{e^{-2 x}}\, d x\)
    \(=\displaystyle \int \frac{e^x-2 e^{-x}}{4 e^{-2 x}+8+e^{2 x}}\, d x\)
    \(=\displaystyle \int \frac{e^x-2 e^{-x}}{4+\left(e^{2 x}+4+4 e^{-2 x}\right)}\, d x\)
    \(=\displaystyle \int \frac{1}{4+u^2}\, d u\)
    \(=\displaystyle \frac{1}{2} \tan ^{-1}\left(\frac{u}{2}\right)+c\)
    \(=\displaystyle \frac{1}{2} \tan ^{-1}\left(\frac{e^x+2 e^{-x}}{2}\right)+c\)

Calculus, EXT1 C2 2020 HSC 13a

  1. Find  `d/(d theta) (sin^3 theta)`.  (1 mark)

  2. Use the substitution  `x = tan theta`  to evaluate  `int_0^1 (x^2)/(1 + x^2)^(5/2)\ dx`.  (4 marks)

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  1. `3 cos theta sin^2 theta`
  2. `sqrt2/12`
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i.   `d/(d theta) (sin^3 theta) = 3 cos theta sin^2 theta`

 

ii.   `text(Let)\ x = tan theta`

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

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

`text(When)\ x = 0, \ theta = 0`

`int_0^1 (x^2)/(1 + x^2)^(5/2) dx` `= int_0^(pi/4) (tan^2 theta)/((1 + tan^2 theta)^(5/2)) xx sec^2 theta\ d theta`
  `= int_0^(pi/4) (tan^2 theta)/((sec^2 theta)^(5/2)) xx sec^2 theta\ d theta`
  `= int_0^(pi/4) (sin^2 theta)/(cos^2 theta) · 1/((sec^2 theta)^(3/2))\ d theta`
  `= int_0^(pi/4) (sin^2 theta)/(cos^2 theta) · 1/(sec^3 theta)\ d theta`
  `= int_0^(pi/4) sin^2 theta cos theta\ d theta`
  `= 1/3[sin^3 theta]_0^(pi/4)`
  `= 1/3(sin^3\ pi/4 – 0)`
  `= 1/3 (1/sqrt2)^3`
  `= 1/(6sqrt2)`
  `= sqrt2/12`

Calculus, EXT1 C2 2019 HSC 13a

Use the substitution  `u = cos^2 x`  to evaluate  `int_0^(pi/4) (sin 2x)/(4 + cos^2 x)\ dx`.  (3 marks)

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`ln {:10/9`

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`u` `= cos^2 x`
`(du)/(dx)` `= -2 sin x cos x`
  `= -sin 2x`
`du` `= -sin 2x\ dx`

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

`text(When)\ \ x = 0,\ \ u = 1`

`:. int_0^(pi/4) (sin 2x)/(4 + cos^2 x)\ dx` `= -int_1^(1/2) (du)/(4 + u)`
  `= -[ln (4 + u)]_1^(1/2)`
  `= -(ln 4.5 – ln 5)`
  `= -ln {:9/10`
  `= ln {:10/9`

Calculus, EXT1 C2 2013 HSC 11f

Use the substitution  `u = e^(3x)`  to evaluate  `int_0^(1/3) (e^(3x))/(e^(6x) + 1)\ dx`.   (3 marks)

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`1/3 (tan^(-1)e\ – pi/4)`

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 MARKER’S COMMENT: Many students did not calculate in radians and incorrectly got an answer of 8.2. BE CAREFUL!
Note that converting your answer to 0.14 is also correct but not required.

`text(Let)\ \ u = e^(3x)`

`(du)/(dx)` `= 3e^(3x)`
`:.dx` `= (du)/(3e^(3x))`

 

`text(When)` `\ x = 1/3,` `\ u = e^(3 xx 1/3) = e`
  `\ x = 0,` `\ u = e^0 = 1`

 
`:.int_0^(1/3) (e^(3x))/(e^(6x) + 1)\ dx`

`=int_1^e (e^(3x))/(u^2 + 1) xx (du)/(3e^(3x))`

`= 1/3 int_1^e 1/(u^2 + 1)\ du`

`= 1/3 [tan^(-1)u]_1^e`

`= 1/3 [tan^(-1) e\ – tan^(-1) 1]`

`= 1/3 (tan^(-1)e\ – pi/4)`