Question

How do I evaluate `int(csc^2x)/(cot^3x) dx`?

Answer 1

You can write:
`int1/sin^2(x)*sin^3(x)/cos^3(x)dx=intsin(x)/cos^3(x)dx=`
You can now use the fact that: `d[cos(x)]=-sin(x)dx`
Your integral becomes:
`int-(d[cos(x)])/cos^3(x)=int-cos^(-3)(x)d[cos(x)]=1/(2*cos^2(x))+c`

Where you used `cos(x)` as if it was `x` in a normal integral integrating `cos^(-3)(x)` as if it was `x^(-3)`

Answer 2

You can use the identity:

`csc^2x = 1 + cot^2x`

If you don't remember that, you can derive it like so:

`sin^2x + cos^2x = 1`

`sin^2x = 1-cos^2x`

`1/sin^2x = 1/(1-cos^2x)`

`color(green)(csc^2x) = (sin^2x + cos^2x)/(1-cos^2x)`

`= (sin^2x + cos^2x)/(sin^2x)`

`= 1 + (cos^2x)/(sin^2x)`

`= color(green)(1 + cot^2x)`

Anyways, you get:

`int csc^2x/(cot^3x)dx`

`= int (1+cot^2x)/(cot^3x)dx`

`= int tan^3x + tanx dx`

`= int (tanx)(tan^2x + 1) dx`

EGADS! Another identity!

`= int tanxsec^2x dx`

Now we can just do a bit of quick u-substitution. Let:

`u = tanx`
`du = sec^2xdx`

`=> int udu = u^2/2 + C`

`= color(blue)(tan^2x/2 + C)`

Now the weird part is, up until and including the `u^2/2`, it seems to be correct. But Wolfram Alpha says it is `sec^2x/2 + C`. If I had to guess, I would say it was because:

`tan^2x/2 + C prop tan^2x/2 + 1/2 + C prop sec^2x/2 + C`

along with a domain/constraints concern.

Answer 3

Here is yet a third method of evaluation.

Explanation:

`int csc^2x/cot^3x dx = int(cotx)^-3 csc^2x dx`

Let `u = cotx` so that `du = -csc^2x dx` and the integral becomes:

`-int u^-3 du = -u^-2 /(-2) +C`

`= u^-2 /2 +C`

`= (cotx)^-2/2 +C`

`= 1/(2cot^2x) +C = tan^2x/2 + C`

Answer 4

And a fourth.

Explanation:

`csc^2x/cot^3x = tan^3x csc^2x`

`= sin^3x/cos^3x 1/sin^2x`

`= sinx/cos^3x`

`= (cosx)^-3 sinx`

So the integral becomes:

`int (cosx)^-3 sinx dx`

which can be evaluated using `u = cosx` to give:

`-(cosx)^-2/-2 +C`

`= sec^2x/2 +C`