# Evaluate the following intcsc^6udu

Question
Trigonometry
Evaluate the following
$$\displaystyle\int{{\csc}^{{6}}{u}}{d}{u}$$

2021-03-10
The trigonometric function $$\displaystyle\int{{\csc}^{{6}}{u}}{d}{u}$$
$$\displaystyle\int{{\csc}^{{6}}{u}}{d}{u}=\int{{\csc}^{{4}}{u}}\cdot{{\csc}^{{2}}{u}}{d}{u}$$
$$\displaystyle=\int{\left({{\csc}^{{2}}{u}}\right)}^{{2}}\cdot{{\csc}^{{2}}{u}}{d}{u}$$
$$\displaystyle=\int{\left({1}+{{\cot}^{{2}}{u}}\right)}^{{2}}\cdot{{\csc}^{{2}}{u}}{d}{u}$$
(Since, $$\displaystyle{{\csc}^{{2}}{u}}={1}+{{\cot}^{{2}}{u}}$$)
Let $$\displaystyle{t}={\cot{{u}}}$$
$$\displaystyle{\frac{{{\left.{d}{t}\right.}}}{{{d}{u}}}}=-{{\csc}^{{2}}{u}}$$ implies $$\displaystyle-{\left.{d}{t}\right.}={{\csc}^{{2}}{u}}{d}{u}$$
Now, $$\displaystyle\int{{\csc}^{{6}}{d}}{u}=-\int{\left({1}+{t}^{{2}}\right)}{\left.{d}{t}\right.}$$
$$\displaystyle=-\int{\left({1}+{2}{t}^{{2}}+{t}^{{4}}\right)}{\left.{d}{t}\right.}$$
$$\displaystyle=-{\left({t}+{\frac{{{2}{t}^{{3}}}}{{{3}}}}+{\frac{{{t}^{{5}}}}{{{5}}}}\right)}+{C}$$
$$\displaystyle=-{t}-{\frac{{{2}{t}^{{3}}}}{{{3}}}}-{\frac{{{t}^{{5}}}}{{{5}}}}+{C}$$
Here $$\displaystyle{t}={\cot{{u}}}$$
Therefore, $$\displaystyle\int{{\csc}^{{6}}{u}}{d}{u}=-{\cot{{u}}}-{\frac{{{2}}}{{{3}}}}{{\cot}^{{3}}{u}}-{\frac{{{1}}}{{{5}}}}{{\cot}^{{5}}{u}}+{C}$$

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