Let's do the trig part first, then the calculus. First we simplify the integrand.
[tex]\cot^3 x \sin ^3 x=\dfrac{\cos^3 x}{\sin^3 x} \sin^3 x= \cos^3 x[/tex]
The triple angle formula is:
[tex]\cos 3x = 4 \cos^3 x - 3 \cos x[/tex]
or
[tex]\cos^3 x = \frac 1 4 (\cos 3x + 3 \cos x)[/tex]
Now we can integrate:
[tex]\displaystyle \int \cot^3 x \sin^3 x \ dx = \int \cos^3 x\ dx= \int \tfrac 1 4 (\cos 3x + 3 \cos x)\ dx[/tex]
[tex]=\frac 1 4 (\frac 1 3 \sin 3x + 3 \sin x) + C[/tex]
[tex]=\frac 1 {12} \sin 3x + \frac 3 4 \sin x + C[/tex]
