<s>[INST] Evaluate $ \int \frac{3x^{2} (\csc{\sqrt{x^{3}-6}})^{2}}{\sqrt{x^{3}-6}} dx $ $$ \int \frac{3x^{2} (\csc{\sqrt{x^{3}-6}})^{2}}{\sqrt{x^{3}-6}} dx = ? $$

Attempt:
$$ \int \frac{3x^{2} (\csc{\sqrt{x^{3}-6}})^{2}}{\sqrt{x^{3}-6}} dx = \int \frac{3x^{2}}{\sqrt{x^{3}-6} \sin^{2}\sqrt{x^{3}-6}} dx $$
$$ U = \sqrt{x^{3}-6} $$
then the integral is
$$ \int \frac{2U}{\sin^{2} U} dU $$
then using partial integration we will get the above equals:
$$ -2 U \cot U + \ln(\sin U) + C = -2 \sqrt{x^{3}-6} \cot(\sqrt{x^{3}-6}) + \ln (\sin \sqrt{x^{3}-6}) + C$$
But...
$$ \frac{d (-2 \cot(\sqrt{x^{3}-6}))}{dx} = \frac{3x^{2} (\csc{\sqrt{x^{3}-6}})^{2}}{\sqrt{x^{3}-6}} $$
 [/INST] When you substitute $U=\sqrt{x^3-6}$, you made a mistake.
It should be $$\int{\dfrac{2U}{U\sin^2{U}}dU}=\int{2\csc^2{U}dU}$$, so the answer is
$$-2\cot^2{\sqrt{x^3-6}}+C$$
</s>