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

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Answer 1 · 2017-09-19T20:55:12

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

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:

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

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