smc-2564-30-Trig

Calculus, SPEC1 2022 VCAA 9

Given that `f^{\prime}(x)=\frac{\cos (2 x)}{\sin ^3(2 x)}` and `f((pi)/(8))=(3)/(4)`, find `f(x)`. (4 marks)

--- 9 WORK AREA LINES (style=lined) ---

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`f(x)=\frac{-1}{4 \sin ^2(2 x)}+\frac{5}{4}=-\frac{1}{4} \text{cosec}^2(2 x)+\frac{5}{4}`

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`text{Let}\ \ u=sin(2x)\ \ =>\ \ {du}/{dx}=2cos(2x)`

`\int \frac{\cos (2 x)}{\sin ^3(2 x)} d x`	`=\frac{1}{2} \int \frac{d u}{u^3}`
	`= -\frac{1}{4} u^{-2}+c`
	`=-\frac{1}{4} \ xx \frac{1}{\sin ^2(2 x)}+c`

`text{When}\ \ x = pi/8:`

`-\frac{1}{4} \cdot \frac{1}{(\frac{1}{sqrt{2}})^2}+c`	`=\frac{3}{4}`
`-\frac{1}{4} \cdot 2+c`	`=\frac{3}{4}`
`\Rightarrow c`	`=\frac{3}{4}+\frac{2}{4}=5/4`

`:. \ f(x)=\frac{-1}{4 \sin ^2(2 x)}+\frac{5}{4}=-\frac{1}{4} \text{cosec}^2(2 x)+\frac{5}{4}`

♦ Mean mark 50%.

Calculus, SPEC2 2020 VCAA 11 MC

With a suitable substitution `int_(pi/4)^(pi/3) (sec^2(x))/(sec^2(x) - 3 tan(x) + 1)\ dx` can be expressed as

`int_1^(1/sqrt3) (1/(u - 1) - 1/(u - 2))\ du`
`int_1^(sqrt3) (1/(3(u - 3)) - 1/(3u))\ du`
`int_1^(sqrt3) (1/(u - 2) - 1/(u - 1))\ du`
`int_1^(sqrt3) (1/(u - 1) - 1/(u - 2))\ du`
`int_(pi/4)^(pi/3) (1/(3(u - 1)) - 1/(3(u + 2)))\ du`

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`C`

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

`(du)/(dx) = sec^2 (x) \ => \ du = sec^2(x)\ dx`

`text(When)\ `	`x = pi/3,\ u = sqrt3`
	`x = pi/4,\ u = 1`

`int_(pi/4)^(pi/3) (sec^2(x))/(sec^2(x) – 3 tan(x) + 1)\ dx`	`= int_1^(sqrt3)\ 1/(u^2 + 1 – 3u + 1)\ du`
	`= int_1^(sqrt3) 1/(u^2 – 3u + 2)\ du`
	`= int_1^(sqrt3) 1/((u – 2)(u – 1))\ du`
	`= int_1^(sqrt3) 1/(u – 2) – 1/(u – 1)\ du`

`=>C`

Calculus, SPEC2 2012 VCAA 13 MC

Using a suitable substitution, `int_0^(pi/3) sin^3 (x) cos^4 (x)\ dx` can be expressed in terms of `u` as

A. `int_0^(pi/3) (u^6 - u^4)\ du`

B. `int_1^(1/2) (u^6 - u^4)\ du`

C. `int_(1/2)^1 (u^6 - u^4)\ du`

D. `int_0^(sqrt3/2) (u^6 - u^4)\ du`

E. `int_0^(sqrt3/2) (u^4 - u^6)\ du`

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`B`

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

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

`sin^2(x)`	`= 1 – cos^2(x)`
	`= 1 – u^2`

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

`text(When)\ \ x = pi/3, \ u = 1/2`

`int_0^(pi/3) sin^3 (x) cos^4 (x)\ dx`

`=-int_1^(1/2) (1 – u^2)u^4\ du`

`= int_1^(1/2) (u^6 – u^4)\ du`

`=> B`

Calculus, SPEC2 2011 VCAA 15 MC

Using a suitable substitution, the definite integral `int_0^(pi/24)tan(2x)sec^2(2x)\ dx` is equivalent to

A. `1/2int_0^(pi/24)(u)\ du`

B. `2int_0^(pi/24)(u)\ du`

C. `int_0^(2 - sqrt3)(u)\ du`

D. `1/2int_0^(2 - sqrt3)(u)\ du`

E. `2int_0^(2 - sqrt3)(u)\ du`

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`D`

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`text(Let)\ \ u = tan(2x)`

`(du)/dx = 2 sec^2(2x)\ \ =>\ \ 1/2 du = sec^2(2x)\ dx`

`text(When)\ \ x=pi/24\ \ =>\ \ u=tan(pi/12)=2-sqrt3\ \ text{(by CAS)}`

`text(When)\ \ x=0\ \ =>\ \ u=0`

`:. int_0^(pi/24)tan(2x)sec^2(2x)\ dx`

`= 1/2int_0^(2 – sqrt3)(u)\ du`

`=> D`

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)`

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`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

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)
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)
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)

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`48`
`int_(sqrt 3/2)^0 – 8u^2 du`
`sqrt 3`

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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`

Calculus, SPEC1 2017 VCAA 6

Let `f(x) = 1/(arcsin(x))`.

Find `fprime(x)` and state the largest set of values of `x` for which `fprime(x)` is defined. (3 marks)

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`x ∈ (−1, 1) \\ {0}`

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`u = sin^(−1)(x), \ f(u(x)) = 1/u`

♦ Mean mark 41%.

`fprime(u(x))`	`= −u^(−2)`
`uprime(x)`	`= 1/sqrt(1 – x^2)`

`fprime(x)`	`= (−u^(−2))/sqrt(1 – x^2)`
`fprime(x)`	`= (−1)/(arcsin^2(x)sqrt(1 – x^2))`

`text(Domain:)\ sin^(−1)(x)\ \ x ∈[−1,1]`

`sin^(−1)(x) != 0 \ \ => \ \ x != 0`

`=> x ∈ [−1, 1] \\ {0}`

`fprime(x)\ text(doesn’t exist at end points)`

`:. x ∈ (−1, 1) \\ {0}`

Calculus, SPEC2 2018 VCAA 8 MC

Using a suitable substitution, `int_0^(pi/6) tan^2 (x) sec^2(x)\ dx` can be expressed as

A. `int_0^(1/sqrt 3) (u^4 + u^2) du`

B. `int_1^(2/sqrt 3) (u^4 + u^2) du`

C. `int_0^(1/(sqrt 3)) u\ du`

D. `int_0^(pi/6) u^2\ du`

E. `int_0^(1/sqrt 3) u^2\ du`

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`E`

Show Worked Solution

`text(Let)\ \ u`	`= tan(x),`
`(du)/(dx)`	`= sec^2 (x)\ \ =>\ du = sec^2(x)\ dx`
`u(0)`	`= tan(0) = 0`
`u (pi/6)`	`= tan (pi/6) = 1/sqrt 3`

`int_0^(pi/6) tan^2 (x)\ sec^2 (x)\ dx = int_0^(1/sqrt 3) u^2 du`

`=> E`