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Calculus, SPEC2-NHT 2017 VCAA 7 MC

`int_(pi/4)^(pi/3) (sin^3(x) cos^2(x))\ dx`  is equivalent to

A.   `int_(1/2)^(1/sqrt 2) (u^4 - u^2)\ du`  where  `u = cos(x)`

B.   `-int_(1/sqrt 2)^(1/2) (u^2 - u^4)\ du`  where  `u = cos(x)`

C.   `-int_(pi/4)^(pi/3) (u^2 - u^4)\ du`  where  `u = sin(x)`

D.   `int_(pi/4)^(pi/3) (u^2 - u^4)\ du`  where  `u = sin(x)`

E.   `- int_(1/sqrt 2)^(1/2) (u^2 - u^4)\ du`  where  `u = sin(x)`

Show Answers Only

`B`

Show Worked Solution

`text(Let)\ \ u = cos (x)`

`sin^3(x)` `= sin(x) ⋅ sin^2(x)`
  `= sin(x) (1 – cos^2(x))`

`(du)/dx = – sin(x)\ \ =>\ \ du = – sin(x)\ dx`

  
`u(pi/3) = 1/2,\ \ u(pi/4) = 1/sqrt 2`
 

`:. int_(pi/4)^(pi/3) sin(x)(1-cos^2(x)) cos^2(x)\ dx`

`=int_(1/sqrt 2)^(1/2) -(1 – u^2)u^2\ du`

`= -int_(1/sqrt 2)^(1/2) (u^2 – u^4)\ du`

 
`=>   B`

Calculus, SPEC1 2014 VCAA 5

  1. For the function with rule  `f(x) = 96 cos (3x) sin (3x)`, Find the value of `a` such that  `f(x) = a sin (6x)`.  (1 mark)
  2. Use an appropriate substitution in the form  `u = g(x)`  to find an equivalent definite integral for

     

         `int_(pi/36)^(pi/12) 96 cos (3x) sin (3x) cos^2 (6x)\ dx`  in terms of `u` only.  (3 marks)

  3. Hence evaluate  `int_(pi/36)^(pi/12) 96 cos (3x) sin (3x) cos^2 (6x)\ dx`, giving your answer in the form  `sqrt k, \ k in Z`.  (1 mark)
Show Answers Only
  1. `48`
  2. `int_(sqrt 3/2)^0 – 8u^2 du`
  3. `sqrt 3`
Show Worked Solution

a.   `96 cos (3x) sin(3x)`

`= 48 (2 cos(3x) sin(3x))`

`= 48 sin (6x)`

`:. a = 48`

 

b. `int_(pi/36)^(pi/12) 96 cos (3x) sin (3x) cos^2 (6x)\ dx`

`= int_(pi/36)^(pi/12) 48 sin (6x) cos^2 (6x)\ dx`

`u` `= cos (6x)`
`(du)/(dx)` `= -6 sin (6x)\ \ =>\ \ du = -6 sin(6x)\ dx`
   
`u(pi/12)` `= cos (pi/2) = 0`
`u(pi/36)` `= cos (pi/6) = sqrt 3/2`

 
`:. int_(pi/36)^(pi/12) 48 sin (6x) cos^2 (6x)\ dx`

`= int_(sqrt 3/2)^0 – 8u^2\ du`

c. `:. int_(pi/36)^(pi/12) 96 cos (3x) sin (3x) cos^2 (6x)\ dx`

`= -8 [u^3/3]_(sqrt 3/2)^0`
`= -8 (0 – (sqrt 3/2)^3/3)`
`= 8/3 xx (3sqrt3)/8`
`= sqrt 3`