Answer
$$\int\cot^3t\csc^4tdt=-\frac{\cot^6t}{6}-\frac{\cot^4t}{4}+C$$
Work Step by Step
$$A=\int\cot^3t\csc^4tdt=\int\cot^3t\csc^2t(\csc^2tdt)$$ $$A=\int\cot^3t(\cot^2t+1)(\csc^2tdt)$$ $$A=-\int\cot^3t(\cot^2t+1)d(\cot t)$$
We set $u=\cot t$ $$A=-\int u^3(u^2+1)du$$ $$A=-\int(u^5+u^3)du$$ $$A=-\frac{u^6}{6}-\frac{u^4}{4}+C$$ $$A=-\frac{\cot^6t}{6}-\frac{\cot^4t}{4}+C$$
