Question

Evaluate the integral? : `int cscx dx`

Answer 1

`int \ cscx \ dx = ln|cscx-cotx| + C`

Explanation:

To derive the result for this integral we multiply numerator and denominator by `cscx-cotx`. This "trick" is a standard technique for this particular result.

We can write the integral as:

`int \ cscx \ dx = int \ cscx \ (cscx-cotx)/(cscx-cotx) \ dx`

`" " = int \ (csc^2x-cotxcscx)/(cscx-cotx) \ dx`

Now we can perform a substitution, Let:

`u = cscx-cotx => (du)/dx = csc^2x-cotxcscx`

Substituting into the integral we get:

`int \ cscx \ dx = int \ 1/u \ du`

`" " = ln|u| + C`

Restoring the substitution we get:

`int \ cscx \ dx = ln|cscx-cotx| " "` QED