Indefinite Integral (09)

[jacks]

Evaluation of \(\displaystyle \int\frac{\cot^2 x}{1+\tan^3 x}dx\)

[Grigorios Kostakos]

A straightforward calculation. I'll be glad to see something smarter! \begin{align*}
\displaystyle \int\frac{\cot^2 x}{1+\tan^3 x}dx&=\int\frac{1}{\tan^2{x}\,(1+\tan{x})\,(1-\tan{x}+\tan^2 x)}dx\\
&\mathop{=\!=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{u\,=\,\tan{x}}\\
{\frac{1}{1+u^2}\,du\,=\,dx}\\
\end{subarray}}\,\int{\frac{1}{u^2\,(1+u)\,(1-u+u^2)}\,\frac{1}{1+u^2}\,du}\\
&=\int{\frac{du}{u^2}}-\frac{1}{2}\int{\frac{u+1}{1+u^2}\,du}+\frac{1}{6}\int{\frac{du}{u+1}}+\frac{1}{3}\int{\frac{u-2}{1-u+u^2}\,du}\\
&=-\frac{1}{u^2}+c_1-\frac{1}{2}\int{\frac{u}{1+u^2}\,du}-\frac{1}{2}\int{\frac{1}{1+u^2}\,du}+\frac{1}{6}\log|u+1|+c_2\,+\\
&\qquad\qquad\frac{1}{3}\int{\Bigl({\frac{1}{2}\,\frac{2u-1}{1-u+u^2}-\frac{3}{2}\,\frac{1}{1-u+u^2}}\Bigr)\,du}\\
&=-\frac{1}{u}+\frac{1}{6}\log|u+1|+c_3-\frac{1}{4}\int{\frac{(1+u^2)'}{1+u^2}\,du}-\frac{1}{2}\,\arctan{u}+c_4\,+\\
&\qquad\qquad\frac{1}{6}\int{\frac{2u-1}{1-u+u^2}\,du}-\frac{1}{2}\int{\frac{1}{1-u+u^2}\,du}\\
&=-\frac{1}{u}+\frac{1}{6}\log|u+1|-\frac{1}{2}\,\arctan{u}-\frac{1}{4}\log({1+u^2})+c_5\,+\\
&\qquad\qquad\frac{1}{6}\int{\frac{(1-u+u^2)'}{1-u+u^2}\,du}-\frac{1}{2}\int{\frac{4}{(1+2u)^2+3}\,du}\\
&=-\frac{1}{u}+\frac{1}{6}\log|u+1|-\frac{1}{2}\,\arctan{u}-\frac{1}{4}\log({1+u^2})+\frac{1}{6}\log({1-u+u^2})+c_6\,-\\
&\qquad\qquad\frac{1}{\sqrt{3}}\int{\frac{1}{\bigl({\frac{1-2u}{\sqrt{3}}}\bigr)^2+1}\,d\bigl({\tfrac{1-2u}{\sqrt{3}}}\bigr)}\\
&=-\frac{1}{u}+\frac{1}{6}\log|u^3+1|-\frac{1}{2}\,\arctan{u}-\frac{1}{4}\log({1+u^2})-\frac{1}{\sqrt{3}}\arctan\bigl({\tfrac{1-2u}{\sqrt{3}}}\bigr)+c\\
&\stackrel{u\,=\,\tan{x}}{=\!=\!=\!=\!=}-\frac{1}{\tan{x}}+\frac{1}{6}\log|\tan^3{x}+1|-\frac{1}{2}\,x-\frac{1}{4}\log({1+\tan^2{x}})-\frac{1}{\sqrt{3}}\arctan\bigl({\tfrac{1-2\tan{x}}{\sqrt{3}}}\bigr)+c\,.
\end{align*}