$\int \sin^3(3x)\cos^{-2}(3x)dx$
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)