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**Soroban** Hello, Preston019!

Did you understand your instructor's example?

If not, no wonder you can't solve this one.

Maybe I'd better baby-step through it . . .

We have: .$\displaystyle I \;=\;\int\csc^3\!x\,dx \;=\;\int \csc x(\csc^2\!x\,dx)$

Integrate by parts: .$\displaystyle \begin{Bmatrix}u &=& \csc x && dv &=& \csc^2\!x\,dx \\ du &=& \text{-}\csc x\cot x\,dx && v &=& \text{-}\cot x \end{Bmatrix}$

We have: .$\displaystyle I \;=\;\text{-}\csc x\cot x - \int\csc x\cot^2\!x\,dx$

. . . . . . . . $\displaystyle I \;=\;\text{-}\csc x\cot x - \int\csc x(\csc^2\!x-1)\,dx$

. . . . . . . . $\displaystyle I \;=\;\text{-}\csc x\cot x - \int(\csc^3\!x - \csc x)\,dx$

. . . . . . . . $\displaystyle I \;=\;\text{-}\csc x\cot x - \underbrace{\int\csc^3\!x\,dx}_{\text{This is }I} + \int\csc x\,dx$

. . . . . . . . $\displaystyle I \;=\;\text{-}\csc x\cot x - I + \int\csc x\,dx$

. . . . . . . $\displaystyle 2I \;=\;\text{-}\csc x\cot x + \int\csc x\,dx $

. . . . . . . $\displaystyle 2I \;=\;\text{-}\csc x\cot x + \ln|\csc x - \cot x| + C$

. . . . . . . . $\displaystyle I \;=\;\tfrac{1}{2}\big(\text{-}\csc x\cot x + \ln|\csc x - \cot x|\big) + C$

Therefore: .$\displaystyle \int\csc^3x\,dx \;=\;\tfrac{1}{2}\big(\text{-}\csc x\cot x + \ln|\csc x - \cot x|\big) + C$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We have: .$\displaystyle I \;=\; \int\!\csc^5\!x\,dx \;=\;\int\!\csc^3\!x(\csc^2\!x\,dx)$

Integrate by parts: .$\displaystyle \begin{Bmatrix}u &=& \csc^3\!x && dv &=& \csc^2\!x\,dx \\ du &=& \text{-}3\csc^3\!x\cot x\,dx && v &=& \text{-}\cot x \end{Bmatrix}$

We have: .$\displaystyle I \;=\;\text{-}\csc^3\!x\cot x - 3\!\int\!\csc^3\!x\cot^2\!x\,dx$

. . . . . . . . $\displaystyle I \;=\;\text{-}\csc^3\!x\cot x - 3\!\int\!\csc^3\!x(\csc^2\!x-1)\,dx$

. . . . . . . . $\displaystyle I \;=\;\text{-}\csc^3\!x\cot x - 3\!\int(\csc^5\!x - \csc^3\!x)\,dx$

. . . . . . . . $\displaystyle I \;=\;\text{-}\csc^3\!x\cot x - 3\underbrace{\int\!\csc^5\!x\,dx}_{\text{This is }I} + 3\!\int\!\csc^3\!x\,dx$

. . . . . . . . $\displaystyle I \;=\;\text{-}\csc^3\!x\cot x - 3I + 3\!\int\!\csc^3\!x\,dx$

. . . . . . .$\displaystyle 4I \;=\;\text{-}\csc^3\!x\cot x + 3\!\int\!\csc^3\!x\,dx$

Since we know the value of $\displaystyle \int\!\csc^3\!x\,dx$, we can finish the problem . . .