Question

How do you find the antiderivative of `int (csc^3x) dx`?

Answer 1

`intcsc^3xdx=(-cscxcotx-lnabs(cotx+cscx))/2`

Explanation:

`intcsc^3xdx=intcsc^2xcscxdx`

Split `csc^3x` into a `cscx` and `csc^2x` term, then use integration by parts. Integration by parts takes the form `intudv=uv-intvdu`. Let:

`{(u=cscx" "=>" "du=-cscxcotxdx),(dv=csc^2x" "=>" "v=-cotx):}`

Thus:

`intcsc^3xdx=-cscxcotx-intcot^2xcscxdx`

Write `cot^2x` as `csc^2x-1`, which arises through the Pythagorean Identity:

`intcsc^3xdx=-cscxcotx-int(csc^2x-1)cscxdx`

`intcsc^3xdx=-cscxcotx-intcsc^3x+intcscxdx`

Integrating `cscx`, which is a common integral (if you don't already know it, a short proof for it is contained at the bottom of this answer):

`intcsc^3xdx=-cscxcotx-intcsc^3xdx-lnabs(cotx+cscx)`

Solving for `intcsc^3xdx` by adding it to both sides:

`2intcsc^3xdx=-cscxcotx-lnabs(cotx+cscx)`

Dividing both sides by `2`:

`intcsc^3xdx=(-cscxcotx-lnabs(cotx+cscx))/2`

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Short proof of the integral of cosecant:

`intcscxdx=intcscx*(cotx+cscx)/(cotx+cscx)dx=int(cscxcotx+csc^2x)/(cotx+cscx)dx`

Let `v=cotx+cscx`, thus `dv=-csc^2x-cscxcotx`, which is the opposite of the numerator. Thus

`intcscxdx=-int(dv)/v=-lnabsv+C=-lnabs(cotx+cscx)+C`