Evaluate the integral? : # int 1/sinx dx#

1 Answer
Sep 19, 2017

# int \ 1/sinx \ dx = -ln | cscx + cotx| + c#

Explanation:

This is a standard result:

# int \ 1/sinx \ dx = int \ csc x \ dx#
# " " = -ln | cscx + cotx| + c#

We can readily derive this result as follows (and having a little knowledge of the expected solution provides a hint):

# int \ 1/sinx \ dx = int \ csc x \ dx#
# " " = int \ csc x * (csc x + cot x)/(csc x + cot x) \ dx # ..... [A]

Now we perform a substitution.

Let #u = csc x + cot x => (du)/dx = -csc x cot x - csc^2 x #

Substituting into the integral [A], we get:

# int \ 1/sinx \ dx = int \ (csc^2x + cscxcot x)/(csc x + cot x) \ dx #
# " " = int \ -1/u \ du #
# " " = - ln |u| + C #

And, restoration of the substitution give us:

# int \ 1/sinx \ dx = - ln |csc x + cot x| + C #