Question

How do you integrate `csc^3x`?

Answer 1

`(-cotxcscx-ln(abs(cotx+cscx)))/2+C`

Explanation:

We have:

`I=intcsc^3xdx`

We will use integration by parts. First, rewrite the integral as:

`I=intcsc^2xcscxdx`

Since integration by parts takes the form `intudv=uv-intvdu`, let:

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

Applying integration by parts:

`I=-cotxcscx-intcot^2xcscxdx`

Through the Pythagorean identity, write `cot^2x` as `csc^2x-1`.

`I=-cotxcscx-int(csc^2x-1)(cscx)dx`

`I=-cotxcscx-intcsc^3xdx+intcscxdx`

Note that `I=intcsc^3xdx` and `intcscxdx=-ln(abs(cotx+cscx))`.

`I=-cotxcscx-I-ln(abs(cotx+cscx))`

Add the original integral `I` to both sides.

`2I=-cotxcscx-ln(abs(cotx+cscx))`

Solve for `I` and add the constant of integration:

`I=(-cotxcscx-ln(abs(cotx+cscx)))/2+C`