Question

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

Answer 1

`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`