Find $\int \sin^4 3x\, dx$ .

-\frac{1}{4}\sin^3 3x\cos 3x+\frac{3x}{8}-\frac{1}{16}\sin 6x+c

\frac{1}{12}\sin^3 3x\cos 3x+\frac{3x}{8}-\frac{1}{16}\sin 6x+c

-\frac{1}{12}\sin^3 3x\cos 3x+\frac{3x}{8}-\frac{1}{16}\sin 6x+c

-\frac{1}{12}\sin^3 3x\cos 3x+\frac{3x}{8}-\frac{1}{12}\sin 6x+c

Explanation

Use

I_n=-\frac{1}{na}\sin^{n-1}ax\cos ax+\frac{n-1}{n}I_{n-2}

with

n=4

and

a=3

I_4=-\frac{1}{12}\sin^3 3x\cos 3x+\frac34 I_2

. Now

I_2=\int\sin^2 3x\,dx=\frac{x}{2}-\frac{\sin6x}{12}

, so

\frac34 I_2=\frac{3x}{8}-\frac{\sin6x}{16}

. Adding gives

-\frac{1}{12}\sin^3 3x\cos 3x+\frac{3x}{8}-\frac{1}{16}\sin6x+c

. The

\frac34

must multiply both parts of

I_2

, not the

x

term alone.

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

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