Find $\int \sin^2 3x\, dx$ .

Model answer

x/2 - sin(6x)/12 + c

Also accepted: x/2-sin6x/12, x/2 - sin 6x/12, x/2 - (1/12)sin(6x) + c, x/2 - (1/12)sin 6x + c, x/2 - (1/12)sin6x, (1/2)x - (1/12)sin(6x) + c

Explanation

With

\theta=3x

\sin^2 3x=\dfrac{1-\cos6x}{2}

. Integrating term by term,

\int\sin^2 3x\,dx=\dfrac{x}{2}-\dfrac{\sin6x}{12}+c

, since

\int\cos6x\,dx=\dfrac{\sin6x}{6}

and that is halved.

Derived from ZIMSEC Further Mathematics 9187/1, November 2010, Q7

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