<s>[INST] Trigonometric integral involving trig multiplication $$\int \sin^3(3x)\cos^{-2}(3x)dx$$
Let$u=3x$; then $du=3dx$, so $dx=\dfrac{du}{3}$
$$\dfrac{1}{3}\int \sin^3(u)\cos^{-2}(u)du$$
Expand $\sin^3(u)$ to $\sin^2(u)\sin(u)$
$$\dfrac{1}{3}\int \sin^2(u)\sin(u)\cos^{-2}(u)du$$
Re-write $\sin^2(u)$ to $(1-\cos^2(u))$
$$\dfrac{1}{3}\int (1-\cos^2(u))\cos^{-2}(u)\sin(u)du$$
Multiply $(1-\cos^2(u))$ by $\cos^{-2}(u)$
$$\dfrac{1}{3}\int (\cos^{-2}(u)-1)\sin(u)du$$
Let $v=\cos(u)$; then $dv=-\sin(u)du$, so $-dv=\sin(u)du$
$$-\dfrac{1}{3}\int (v^{-2}-1)dv$$
Integrate
$$-\dfrac{1}{3}(-v^{-1}-v)$$
Re-write in terms of $x$
$$-\dfrac{1}{3}(-\cos^{-1}(3x)-\cos(3x))+C$$
Incorrect answer according to MyMathLab (Pearson Education)
 [/INST] Your answer is correct.  However, note that 
$$-\dfrac{1}{3}(-\frac{1}{\cos(3x)}-\cos(3x)) = \dfrac{1}{3} (\cos(3x)+\sec(3x))$$
Differentiating your answer, we have:
$$\dfrac{1}{3} \frac{\text{d}}{\text{d}x} (\cos(3x)+\sec(3x))+C = \dfrac{1}{3} \cdot 3 (-\sin (3x)+\tan (3x) \sec (3x)) = \sin(3x)\tan^2(3x) = \frac{\sin^3 (3x)}{\cos^2 (3x)}$$
So your integral is correct.

Try and avoid the $\sin^{-1}$ notation for $\frac{1}{\sin x}$, as people may confuse with $\arcsin$.
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