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Calculus, 2ADV C4 2019 HSC 9 MC

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

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

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`text(Consider option)\ A:`

♦♦ Mean mark 35%.

`d/(dx) (tan x – x + C)`

`= sec^2 x – 1`

`= tan^2 x`

`:. int tan^2 x\ dx = tan x – x + C`

`=>  A`

Calculus, 2ADV C4 2010 HSC 5b

  1. Prove that  `sec^2 x + secx tanx = (1 + sinx)/(cos^2x)`.   (1 mark)

  2. Hence prove that  `sec^2 x + secx tanx = 1/(1 - sinx)`.     (1 mark)

  3. Hence, use the identity  `int sec ax tan ax\ dx=1/a sec ax`  to find the exact value of

     

          `int_0^(pi/4) 1/(1 - sinx)\ dx`.   (2 marks)

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  1. `text(Proof)  text{(See Worked Solutions)}`
  2. `text(Proof)  text{(See Worked Solutions)}`
  3. `sqrt2`
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i.    `text(Need to prove)`

`sec^2x + secxtanx = (1 + sinx)/(cos^2x)`
 

`text(LHS)` `=sec^2x + secx tanx`
  `=1/(cos^2x) + 1/(cosx) xx (sinx)/cosx`
  `=1/(cos^2x) + (sinx)/(cos^2x)`
  `=(1 + sinx)/(cos^2x)`
  `= text(RHS)\ \ \ \ text(… as required)`

 

ii.   `text(Need to prove)` 

♦♦ Mean mark 31%.
`sec^2x + secx tanx` `= 1/(1\ – sinx)`
`text(i.e.)\ \ (1 + sinx)/(cos^2x)` `= 1/(1\ – sin x)\ \ \ \ \ text{(part (i))}`
`text(LHS)` `= (1 + sinx)/(cos^2x)`
  `=(1 + sin x)/(1\ – sin^2x)`
  `=(1 + sinx)/((1\ – sinx)(1 + sinx)`
  `=1/(1\ – sinx)\ \ \ \ text(… as required)`

 

iii.  `int_0^(pi/4) 1/(1\ – sinx)\ dx`

♦ Mean mark 37%.

`= int_0^(pi/4) (sec^2x + secx tanx)\ dx`

`= [tanx + secx]_0^(pi/4)`

`= [(tan(pi/4) + sec(pi/4)) – (tan0 + sec0)]`

`= [(1 + 1/(cos(pi/4)))\ – (0 + 1/(cos0))]`

`= 1 + sqrt2\ – 1`

`= sqrt2`