Question

$\int \frac{\cot x+{\cot }^{3}x}{1+{\cot }^{3}x}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \left(\frac{\cot x+{\cot }^{3}x}{1+{\cot }^{3}x}\right)dx \\ =\int \left[\frac{\cot x\left(1+{\cot }^{2}x\right)}{1+{\cot }^{3}x}\right]dx \\ =\int \left(\frac{\cot x{\mathrm{cosec}}^{2}x}{1+{\cot }^{3}x}\right)dx \\ \mathrm{Putting}\cot x=t \\ \Rightarrow -{\mathrm{cosec}}^{2}xdx=dt \\ \Rightarrow {\mathrm{cosec}}^{2}xdx=-dt \\ \therefore I=-\int \frac{tdt}{1+t^3} \\ =-\int \frac{tdt}{\left(1+t\right)\left(t^2-t+1\right)} \\ \mathrm{Let}\frac{t}{\left(1+t\right)\left(t^2-t+1\right)}=\frac{A}{t+1}+\frac{Bt+C}{t^2-t+1} \\ \Rightarrow \frac{t}{\left(1+t\right)\left(t^2-t+1\right)}=\frac{A\left(t^2-t+1\right)+\left(Bt+C\right)\left(t+1\right)}{\left(t+1\right)\left(t^2-t+1\right)} \\ \Rightarrow t=A\left(t^2-t+1\right)+B{t}^2+Bt+Ct+C \\ \Rightarrow t=\left(A+B\right)t^2+\left(B+C-A\right)t+A+C \\ \mathrm{Equating}\mathrm{Coefficients}\mathrm{of}\mathrm{like}\mathrm{terms} \\ A+B=0.....\left(1\right) \\ B+C-A=1.....\left(2\right) \\ A+C=0.....\left(3\right) \\ \mathrm{Solving}\left(1\right),\left(2\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get} \\ A=-\frac{1}{3} \\ B=\frac{1}{3} \\ C=\frac{1}{3} \\ \therefore \frac{t}{\left(1+t\right)\left(t^2-t+1\right)}=-\frac{1}{3\left(t+1\right)}+\frac{1}{3}\left(\frac{t+1}{t^2-t+1}\right) \\ \Rightarrow \frac{t}{\left(1+t\right)\left(t^2-t+1\right)}=-\frac{1}{3\left(t+1\right)}+\frac{1}{6}\left[\frac{2t+2}{t^2-t+1}\right] \\ \Rightarrow \frac{t}{\left(1+t\right)\left(t^2-t+1\right)}=-\frac{1}{3\left(t+1\right)}+\frac{1}{6}\left[\frac{2t-1+3}{t^2-t+1}\right] \\ \therefore I=-\left[-\frac{1}{3}\int \frac{dt}{t+1}+\frac{1}{6}\int \left(\frac{2t-1}{t^2-t+1}\right)dt+\frac{1}{2}\int \frac{dt}{t^2-t+1}\right] \\ =+\frac{1}{3}\int \frac{dt}{t+1}-\frac{1}{6}\int \left(\frac{2t-1}{t^2-t+1}\right)dt-\frac{1}{2}\int \frac{dt}{t^2-t+{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{4}}+1} \\ =\frac{1}{3}\int \frac{dt}{t+1}-\frac{1}{6}\int \frac{\left(2t-1\right)dt}{\left(t^2-t+1\right)}-\frac{1}{2}\int \frac{dt}{{\left(t-{\displaystyle \frac{1}{2}}\right)}^2+{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^2} \\ \mathrm{let}{t}^2-t+1=p \\ \Rightarrow \left(2t-1\right)dt=dp \\ \therefore I=\frac{1}{3}\int \frac{dt}{t+1}-\frac{1}{6}\int \frac{dp}{p}-\frac{1}{2}\int \frac{dt}{{\left(t-{\displaystyle \frac{1}{2}}\right)}^2+{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^2} \\ =\frac{1}{3}\log \left|t+1\right|-\frac{1}{6}\log \left|p\right|-\frac{1}{2}\times \frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{t-{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{\sqrt{3}}{2}}}\right)+C \\ =\frac{1}{3}\log \left|t+1\right|-\frac{1}{6}\log \left|p\right|-\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2t-1}{\sqrt{3}}\right)+C \\ =\frac{1}{3}\log \left|\cot x+1\right|-\frac{1}{6}\log \left|{\cot }^{2}x-\cot x+1\right|-\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2cotx-1}{\sqrt{3}}\right)+C \end{gathered}$$