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Year 12 Maths Advanced (2027) Integral calculus

Areas Involving Trigonometric Functions

20 practice questions 0 video lessons Theory + worked examples

Learn areas involving trigonometric functions for NSW Year 12 Mathematics Advanced (NESA outcome MAV-12-05). Sine and cosine curves rise above and dip below the \(x\)-axis in regular arches, so their areas need integration in radians and careful attention to sign.

You will learn to integrate \(\sin x\) and \(\cos x\) over exact intervals, split regions where a curve crosses the axis, add the separate parts as positive areas and use symmetry to shorten the working — skills used whenever a wave or cycle is measured.

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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.

Note: areas enclosed between two curves are studied in Mathematics Extension 1, not in this course.

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.