Evaluate $ \int \frac{3x^{2} (\csc{\sqrt{x^{3}-6}})^{2}}{\sqrt{x^{3}-6}} dx $

Question

$$ \int \frac{3x^{2} (\csc{\sqrt{x^{3}-6}})^{2}}{\sqrt{x^{3}-6}} dx = ? $$

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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...

$$ \frac{d (-2 \cot(\sqrt{x^{3}-6}))}{dx} = \frac{3x^{2} (\csc{\sqrt{x^{3}-6}})^{2}}{\sqrt{x^{3}-6}} $$

Answer 1

$$u=\sqrt{x^3-6},du=\dfrac{3x^2}{2\sqrt{x^3-6}}dx$$

$$2\int\csc^2u=-2\cot u+K$$

Answer 2

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 $$-2\cot^2{\sqrt{x^3-6}}+C$$