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

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

c.    `sqrt2`

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a.    `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)`

  
b.    `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 (a))}`
`text(LHS)` `= (1 + sinx)/(cos^2x)`
  `=(1 + sin x)/(1-sin^2x)`
  `=(1 + sinx)/((1-sinx)(1 + sinx)`
  `=1/(1-sinx)\ \ \ \ text(… as required)`

  
c.    `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`