<s>[INST] Calculating $\int \dfrac{\cot^3x}{\sqrt{1+\csc^4x}}\;dx$ 
Evaluate:$$\int \dfrac{\cot^3x}{\sqrt{1+\csc^4x}}\;dx$$

 [/INST] Rewrite the integral as
$$\int dx \frac{\cot{x} (\csc^2{x}-1)}{\sqrt{1+\csc^4{x}}}$$
Split the integral along the numerator.  The first piece is
$$\int dx \frac{\csc^2{x} \cot{x}}{\sqrt{1+\csc^4{x}}} = -\int \frac{d(\cot{x}) \cot{x}}{\sqrt{1+(1+\cot^2{x})^2}} = -\frac12 \int \frac{du}{\sqrt{(u+1)^2+1}}$$
where $u=\cot^2{x}$.  The second piece is
$$-\int dx \frac{\cot{x}}{\sqrt{1+\csc^4{x}}} = -\int dx \frac{\cos{x}\sin{x}}{\sqrt{1+\sin^4{x}}} = -\frac12 \int \frac{dv}{\sqrt{1+v^2}}$$
where $v=\sin^2{x}$.  Use
$$\int \frac{dw}{\sqrt{w^2+1}} = \log{(w+\sqrt{w^2+1})}+C$$
and you should be able to finish this off.
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