Harder Trigonometric Calculus

Calculus, EXT1 C2 2025 HSC 11g

Evaluate \(\displaystyle\int_{\small{\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{3}}} \cos ^2(3 x) d x\). (3 marks)

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\(\dfrac{\pi}{12}\)

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\(\displaystyle\int_{\small{\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{3}}} \cos ^2(3x) dx\)	\(=\displaystyle \dfrac{1}{2} \int_{\small{\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{3}}} (\cos6x+1) dx \)
	\(=\dfrac{1}{2}\left[\dfrac{1}{6} \sin 6 x+x\right]_{\small{\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{3}}}\)
	\(=\dfrac{1}{2}\left[\left(\dfrac{1}{6} \sin 2 \pi+\dfrac{\pi}{3}\right)-\left(\dfrac{1}{6} \sin \pi+\dfrac{\pi}{6}\right)\right]\)
	\(=\dfrac{1}{2}\left(\dfrac{\pi}{3}-\dfrac{\pi}{6}\right)\)
	\(=\dfrac{\pi}{12}\)

Calculus, EXT1 C2 2025 HSC 11c

Find \(\displaystyle \int \sin 3x \, \cos x \, dx\). (2 marks)

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\(-\dfrac{1}{8} \cos 4 x-\dfrac{1}{4} \cos 2 x+c\)

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\(\displaystyle\int \sin3x \, \cos x \, dx\)	\(=\displaystyle\frac{1}{2} \int \sin 4 x+\sin2x \,dx\)
	\(=-\dfrac{1}{8} \cos 4 x-\dfrac{1}{4} \cos 2 x+c\)

Calculus, EXT1 C2 2024 HSC 13b

Show that \(\cos ^4 x+\sin ^4 x=\dfrac{1+\cos ^2 2 x}{2}\). (2 marks)

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Hence, or otherwise, evaluate \(\displaystyle{\int}_0^{\frac{\pi}{4}}\left(\cos ^4 x+\sin ^4 x\right) d x\). (3 marks)

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i.	\(\text{LHS}\)	\(=\left[\dfrac{1}{2}(1+\cos (2 x)\right]^2+\left[\dfrac{1}{2}(1-\cos (2 x)\right]^2\)
		\(=\dfrac{1}{4}\left(1+2 \cos (2 x)+\cos ^2(2 x)+1-2 \cos (2 x)+\cos ^2(2 x)\right)\)
		\(=\dfrac{1}{4}\left(2+2 \cos ^{2}(2 x)\right)\)
		\(=\dfrac{1+\cos ^2(2 x)}{2}\)

ii. \(\dfrac{3 \pi}{16}\)

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i.	\(\text{LHS}\)	\(=\left[\dfrac{1}{2}(1+\cos (2 x)\right]^2+\left[\dfrac{1}{2}(1-\cos (2 x)\right]^2\)
		\(=\dfrac{1}{4}\left(1+2 \cos (2 x)+\cos ^2(2 x)+1-2 \cos (2 x)+\cos ^2(2 x)\right)\)
		\(=\dfrac{1}{4}\left(2+2 \cos ^{2}(2 x)\right)\)
		\(=\dfrac{1+\cos ^2(2 x)}{2}\)

ii.	\(\displaystyle{\int}_0^{\frac{\pi}{4}}\left(\cos ^4 x+\sin ^4 x\right) d x\)
		\(=\dfrac{1}{2} \displaystyle{\int}_0^{\frac{\pi}{4}} 1+\cos ^2(2 x) d x\)
		\(=\dfrac{1}{2} \displaystyle{\int}_0^{\frac{\pi}{4}} 1+\dfrac{1}{2}(1+\cos (4 x)) d x\)
		\(=\dfrac{1}{2}\left[\dfrac{3}{2}x +\dfrac{1}{8} \sin (4 x)\right]_0^{\frac{\pi}{4}}\)
		\(=\dfrac{1}{2}\left[\dfrac{3}{2} \times \dfrac{\pi}{4}+\dfrac{1}{8} \sin \pi-0\right]\)
		\(=\dfrac{3 \pi}{16}\)

Calculus, EXT1 C2 2021 HSC 2 MC

Which of the following integrals is equivalent to `int sin^2 3x\ dx`?

`int (1 + cos6x)/2 dx`
`int (1 - cos6x)/2 dx`
`int (1 + sin6x)/2 dx`
`int (1 - sin6x)/2 dx`

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`B`

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`int sin^2 3x\ dx = 1/2 int (1 – cos6x) dx`

`=>\ B`

Calculus, EXT1 C2 2020 HSC 13c

Suppose `f(x) = tan(cos^(-1)(x))` and `g(x) = (sqrt(1-x^2))/x`.

The graph of `y = g(x)` is given.

Show that `f^(′)(x) = g^(′)(x)`. (4 marks)

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Using part (i), or otherwise, show that `f(x) = g(x)`. (3 marks)

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i. `text(See Worked Solutions)`

ii. `text(See Worked Solutions)`

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i. `f(x) = tan(cos^(-1)(x))`

♦ Mean mark (i) 50%.

`f^(′)(x)`	`= -1/sqrt(1-x^2) · sec^2(cos^(-1)(x))`
	`= -1/sqrt(1-x^2) · 1/(cos^2(cos^(-1)(x)))`
	`= -1/(x^2sqrt(1-x^2))`

`g(x) = (1-x^2)^(1/2) · x^(-1)`

`g^(′)(x)`	`= 1/2 · -2x(1-x^2)^(-1/2) · x^(-1)-(1-x^2)^(1/2) · x^(-2)`
	`= (-x)/(x sqrt(1-x^2))-sqrt(1-x^2)/(x^2)`
	`= (-x^2-sqrt(1-x^2) sqrt(1-x^2))/(x^2 sqrt(1-x^2))`
	`= (-x^2-(1-x^2))/(x^2sqrt(1-x^2))`
	`= -1/(x^2sqrt(1-x^2))`
	`=f^(′)(x)`

ii. `f^(′)(x) = g^(′)(x)`

♦♦♦ Mean mark (ii) 15%.

`=> f(x) = g(x) + c`

`text(Find)\ c:`

`f(1)= tan(cos^(-1) 1)= tan 0=0`

`g(1) = sqrt(1-1)/0 = 0`

`f(1) = g(1) + c\ \ =>\ \ c = 0`

`:. f(x) = g(x)`

Calculus, EXT1 C2 2020 HSC 12d

Find `int_0^(pi/2) cos 5x\ sin 3x\ dx`. (3 marks)

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`−1/2`

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`int_0^(pi/2) cos 5x\ sin 3x\ dx`	`= 1/2 int_0^(pi/2) 2cos 5x\ sin 3x\ dx`
	`= 1/2 int_0^(pi/2) sin 8x-sin 2x\ dx`
	`= 1/2[−1/8 cos 8x + 1/2 cos 2x]_0^(pi/2)`
	`= 1/2[(−1/8 cos 4pi + 1/2 cospi)-(−1/8 cos0 + 1/2 cos0)]`
	`= 1/2(−1/8-1/2 + 1/8-1/2)`
	`= −1/2`

Calculus, EXT1 C2 2019 HSC 14c

The diagram shows the two curves `y = sin x` and `y = sin(x-alpha) + k`, where `0 < alpha < pi` and `k > 0`. The two curves have a common tangent at `x_0` where `0 < x_0 < pi/2`.

Explain why `cos x_0 = cos (x_0-alpha)`. (1 mark)

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Show that `sin x_0 = -sin(x_0-alpha)`. (2 marks)

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Hence, or otherwise, find `k` in terms of `alpha`. (2 marks)

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`text(See Worked Solutions)`
`text(Proof)\ text{(See Worked Solutions)}`
`k = 2 sin\ alpha/2`

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i.	`y_1`	`= sin x`
	`(dy_1)/(dx)`	`= cos x`
	`y_2`	`= sin(x-alpha) + k`
	`(dy_2)/(dx)`	`= cos (x-alpha)`

`text(At)\ \ x = x_0,\ \ text(tangent is common)`

♦ Mean mark part (i) 47%.

`:. cos x_0 = cos(x_0-alpha)`

ii. `x_0\ text{is in 1st quadrant (given).}`

`text{Using part (i):}`

`cos\ x_0 = cos(x_0-alpha) >0`

♦♦♦ Mean mark part (ii) 19%.

`=> x_0-alpha\ text(is in 4th quadrant)\ (0 < alpha < pi)`

`text(S)text(ince sin is positive in 1st quadrant and)`

`text(negative in 4th quadrant)`

`=> sin x_0 = -sin(x_0-alpha)`

iii.

`text(When)\ \ x = x_0:`

`y_1`	`=sin x_0`
`y_2`	`=sin(x_0-alpha) + k`
`sin x_0`	`=sin (x_0-alpha) + k`
`sin x_0`	`= -sin x_0 + k`
`k`	`== 2\ sin x_0`

♦♦ Mean mark part (iii) 21%.

`text(S)text(ince)\ \ cos x_0`	`= cos(x_0-alpha)`
`x_0`	`= -(x_0-alpha)`
`2x_0`	`= alpha`
`x_0`	`= alpha/2`

`:. k = 2 sin\ alpha/2`

Calculus, EXT1 C2 2019 HSC 11e

Find `int 2 sin^2 4x\ dx`. (2 marks)

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`x-1/8 sin x + c`

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`text(Using)\ \ sin^2theta = 1/2 (1-cos 2theta):`

`int 2 sin^2 4x\ dx`	`= int 2 xx 1/2 (1-cos 8x)\ dx`
	`= int 1-cos 8x\ dx`
	`= x-1/8 sin 8x + C`

Calculus, EXT1 C2 2016 HSC 5 MC

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

`1/2(x-1/4 sin4x) + c`
`1/2(x + 1/4 sin4x) + c`
`(sin^3 2x)/6 + c`
`(-cos^3 2x)/6 + c`

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`A`

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`int sin^2 2x\ dx`

`= 1/2 int (1-cos 4x)\ dx`

`= 1/2 (x-1/4 sin 4x) + c`

`=> A`

Calculus, EXT1 C2 2005 HSC 3b

By expanding the left-hand side, show that
`qquad sin(5x + 4x) + sin(5x-4x) = 2 sin (5x) cos(4x)` (1 mark)

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Hence find `int sin(5x) cos (4x)\ dx.` (2 marks)

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

b. `-1/18 cos(9x)-1/2 cos(x) + c`

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a. `sin (5x + 4x) + sin (5x-4x) = 2 sin(5x) cos(4x)`

`text(LHS)`	`= sin (5x) cos (4x)-sin(4x) cos (5x) + sin (5x) cos (4x)+ sin (4x) cos (5x)`
	`= 2 sin (5x) cos (4x)\ \ text(… as required)`

b. `int sin (5x) cos (4x)\ dx`

`= 1/2 int 2 sin (5x) cos (4x)\ dx`

`= 1/2 int sin (5x + 4x) + sin (5x-4x)\ dx`

`= 1/2 int sin (9x) + sin (x)\ dx`

`= 1/2 [-1/9 cos(9x)-cos(x)] + c`

`= -1/18 cos(9x)-1/2 cos(x) + c`

Calculus, EXT1 C2 2010 HSC 2a

The derivative of a function `f(x)` is given by

`f^{′}(x) = sin^2 x`.

Find `f(x)`, given that `f(0) = 2`. (2 marks)

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`1/2x-1/4 sin 2x + 2`

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`f^{′}(x)`	`= sin^2 x`
`f(x)`	`= int sin^2 x\ dx`
	`= int 1/2(1-cos 2x)\ dx`
	`= int 1/2-1/2 cos 2x\ dx`
	`= 1/2 x-1/4 sin 2 x + c`

`text(Given)\ \ f(0) = 2,`

`2`	`= 1/2 xx 0-1/4 sin 0 + c`
`:. c= 2`

`:.\ f(x) = 1/2 x-1/4 sin 2x + 2`

Calculus, EXT1 C2 2012 HSC 7 MC

Which expression is equal to `int sin^2 3x\ dx`?

`1/2 (x-1/3 sin 3x) + C`
`1/2 (x + 1/3 sin 3x) + C`
`1/2 (x-1/6 sin 6x) + C`
`1/2 (x + 1/6 sin 6x) + C`

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`C`

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`text(Using:)\ \ sin^2a = 1/2 (1-cos 2a)`

`int sin^2 3x\ dx`	`= 1/2 int (1-cos 6x)\ dx`
	`= 1/2 (x-1/6 sin 6x) + C`

`=> C`