Evaluate the integral? : # int cscx dx#

1 Answer
Jul 4, 2017

# 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