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Year 12 Maths Advanced (2027) Integral calculus

Areas Involving Trigonometric Functions

20 practice questions 0 video lessons Theory + worked examples
NSW · Year 12 Mathematics Advanced · Integral calculus

Areas involving trigonometric functions use definite integrals of \(\sin x\) and \(\cos x\) to find regions under and between trig curves (radians).

Part of the NSW Year 12 Mathematics Advanced course, in the Calculus area of study (Integral calculus focus area) of the 2024 syllabus. Work through practice questions with fully worked solutions and video lessons, or scroll down for the theory summary and worked examples.

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Theory

Areas under trig curves use the trig integral rules (radians). This Year 12 Mathematics Advanced topic (MAV-12-05) evaluates areas under \(\sin\) and \(\cos\).

Area \(=\displaystyle\int_a^b f(x)\,dx\) for \(f\ge0\). \(\sin x\ge0\) on \([0,\pi]\), so its area there is a clean integral. Where the curve dips below the axis, take the size of that part.

Area under a sine curveThe area under y = sin x from 0 to pi is 2. xy y=sin x π
Area under \(y=\sin x\) from \(0\) to \(\pi\) is \(2\).
\[\int_0^{\pi}\sin x\,dx=\big[-\cos x\big]_0^{\pi}=2\]
integral from 0 to pi of sin x equals 2

Method

  1. Integrate the trig function.
  2. Substitute the limits (radians).
  3. Take the size of any part below the axis.
Example 1 — Under sin x
Find the area under \(y=\sin x\) from \(0\) to \(\pi\).
Solution
\(\ \)\(=\)\(\big[-\cos x\big]_0^{\pi}=2\)
Example 2 — Under cos x
Find the area under \(y=\cos x\) from \(0\) to \(\dfrac{\pi}{2}\).
Solution
\(\ \)\(=\)\(\big[\sin x\big]_0^{\pi/2}=1\)
Example 3 — Half a hump
Find the area under \(y=\sin x\) from \(0\) to \(\dfrac{\pi}{2}\).
Solution
\(\ \)\(=\)\(\big[-\cos x\big]_0^{\pi/2}=1\)
Example 4 — Definite
Evaluate \(\displaystyle\int_0^{\pi} \sin x\,dx\).
Solution
\(\ \)\(=\)\(2\)

Common pitfalls

Sine integrates to \(-\cos\) — watch the sign at the limits.
Radians throughout.
Below the axis needs the size of the integral.

Frequently asked questions

What is the area under sin x from 0 to pi?

It is 2, from minus cos x evaluated between 0 and pi.

What is the area under cos x from 0 to pi over 2?

It is 1, from sin x evaluated between 0 and pi over 2.

Why must the angles be in radians?

The trig integral rules only hold when the angle is measured in radians.

What if the sine curve goes below the axis?

Over that part the integral is negative, so take its absolute value for the area.