Question 1(a) Achieved

A sec–tan integral, an area under a cubic-in-a-bracket, a kinematics chain that starts 360 m from the reference point, a trig differential equation, and an area between two trig curves.

Find int 6 sec(2x) tan(2x) dx.

The 2024 paper, page 2 ↗ Marking schedule, page 1 ↗

Work it out

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

1. 1 Which function has sec x tan x as its derivative?

   sec x. So int sec x tan x dx = sec x + c.

2. 2 The angle is 2x. What do you divide by?

   By 2, the number in front of x: int sec(2x) tan(2x) dx = (1/2) sec(2x) + c.

3. 3 Bring in the 6.

   6 × (1/2) sec(2x) + c = 3 sec(2x) + c

4. 4 Check it by differentiating your answer.

   (d/dx)(3 sec(2x)) = 3 × 2 sec(2x) tan(2x) = 6 sec(2x) tan(2x)

What the marker wanted

• Achieved The correct integral, 3 sec 2x. This year’s schedule says the + c is not required for it.

In our words, from NZQA’s 2024 marking schedule.

How did it go?

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