# How to find integral -csc^2x/cotx^1/2 dx?

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### 1 Answer

The integral `int(-csc^2x)/((cotx)^(1/2)) dx` has to be determined.

Make the substitution: `cotx=u`

Then `-csc^2xdx=du`

The given integral takes the form after substitution:

`int((du)/u^(1/2))`

`=int(u^(-1/2)du)`

`=u^(-1/2+1)/(-1/2+1)+C` (where C is the constant of integration)

`=u^(1/2)/(1/2)+C`

`=2u^(1/2)+C`

Put back the value of u to get the final answer as:

`=2(cotx)^(1/2)+C`