Answer
$$\ln \left| {\frac{{\sqrt {1 + 2\sin x} - 1}}{{\sqrt {1 + 2\sin x} + 1}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\cot x}}{{\sqrt {1 + 2\sin x} }}} dx \cr
& {\text{Use the identity }}\cot x = \frac{{\cos x}}{{\sin x}} \cr
& = \int {\frac{{\cos x}}{{\sin x\sqrt {1 + 2\sin x} }}} dx \cr
& {\text{Let }}u = \sin x,{\text{ }}du = \cos xdx \cr
& \int {\frac{{\cos x}}{{\sin x\sqrt {1 + 2\sin x} }}} dx = \int {\frac{{du}}{{u\sqrt {1 + 2u} }}} \cr
& {\text{Use the Table of Integrals on the Reference Pages}} \cr
& {\text{*}}\int {\frac{{du}}{{u\sqrt {a + bu} }} = \frac{1}{{\sqrt a }}\ln \left| {\frac{{\sqrt {a + bu} - \sqrt a }}{{\sqrt {a + bu} + \sqrt a }}} \right| + C,{\text{ }}a > 0,{\text{ then}}} \cr
& \int {\frac{{du}}{{u\sqrt {1 + 2u} }}} = \frac{1}{{\sqrt 1 }}\ln \left| {\frac{{\sqrt {1 + 2u} - \sqrt 1 }}{{\sqrt {1 + 2u} + \sqrt 1 }}} \right| + C \cr
& {\text{Simplify}} \cr
& \int {\frac{{du}}{{u\sqrt {1 + 2u} }}} = \ln \left| {\frac{{\sqrt {1 + 2u} - 1}}{{\sqrt {1 + 2u} + 1}}} \right| + C \cr
& {\text{Write in terms of }}x \cr
& \int {\frac{{\cot x}}{{\sqrt {1 + 2\sin x} }}} dx = \ln \left| {\frac{{\sqrt {1 + 2\sin x} - 1}}{{\sqrt {1 + 2\sin x} + 1}}} \right| + C \cr} $$
