Answer
$$\int 8\cot^4tdt=-\frac{8}{3}\cot^3t+8\cot t+8t+C$$
Work Step by Step
$$A=\int 8\cot^4tdt$$
Use Reduction Formula 94, which states that
$$\int\cot^naxdx=-\frac{\cot^{n-1}ax}{a(n-1)}-\int\cot^{n-2}axdx$$
for $n=4$ and $a=1$
$$A=8\Big[-\frac{\cot^3t}{3}-\int\cot^2tdt\Big]$$
Use Formula 94 again, this time for $n=2$ and $a=1$
$$A=8\Big[-\frac{\cot^3t}{3}-\Big(-\frac{\cot t}{1}-\int\cot^0tdt\Big)\Big]$$ $$A=8\Big[-\frac{\cot^3t}{3}-\Big(-\cot t-\int 1dt\Big)\Big]$$ $$A=8\Big[-\frac{\cot^3t}{3}+\cot t+t\Big]+C$$ $$A=-\frac{8}{3}\cot^3t+8\cot t+8t+C$$
