AS 91579 · 2025 paper · standard integrals, separation of variables

Question 1(a) Achieved

Five independent parts, no shared scenario: a standard integral, a differential equation with a boundary condition, a pair of definite integrals that pin down two constants, a trig area, and a differential equation you have to rearrange before you can separate it.

Find int 6 sec(3x) tan(3x) dx.

The 2025 paper, page 2 ↗ Marking schedule, page 1 ↗

Work it out

Try it on paper first. Stuck? Take one hint at a time.

1 Which function has sec x tan x as its derivative?
(d/dx) sec x = sec x tan x, so int sec x tan x dx = sec x + c.
2 The angle here is 3x, not x. What does the chain rule put in front when you differentiate sec(3x)?
(d/dx) sec(3x) = 3 sec(3x) tan(3x). So the integral of sec(3x) tan(3x) is (1/3) sec(3x): divide by the 3.
3 Now bring in the 6, and finish with the constant.
int 6 sec(3x) tan(3x) dx = 6 × (1/3) sec(3x) + c = 2 sec(3x) + c
4 Check it by differentiating your answer.
(d/dx)(2 sec(3x)) = 2 × 3 sec(3x) tan(3x) = 6 sec(3x) tan(3x), the integrand.

Answer 2 sec(3x) + c

What the marker wanted

Achieved The correct answer, 2 sec 3x + c.

In our words, from NZQA’s 2025 marking schedule.

Common mistake

Leaving off the + c. NZQA’s 2025 report lists ignoring the constant of integration, or treating it as zero, among the things Not Achieved scripts had in common.

How did it go?

Got it moves it further out. Review brings it back tomorrow.