Answer
$$\frac{1}{6}\ln \left( {\frac{{1 - \sin \theta }}{{\sin \theta + 5}}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\cos \theta }}{{{{\sin }^2}\theta + 4\sin \theta - 5}}d\theta } \cr
& {\text{Let }}x = \sin \theta ,{\text{ }}dx = \cos \theta d\theta \cr
& \int {\frac{{\cos \theta }}{{{{\sin }^2}\theta + 4\sin \theta - 5}}d\theta } = \int {\frac{{dx}}{{{x^2} + 4x - 5}}} \cr
& {\text{Decomposing }}\frac{1}{{{x^2} + 4x - 5}}{\text{ into partial fractions}} \cr
& \frac{1}{{{x^2} + 4x - 5}} = \frac{1}{{\left( {x + 5} \right)\left( {x - 1} \right)}} \cr
& \frac{1}{{\left( {x + 5} \right)\left( {x - 1} \right)}} = \frac{A}{{x + 5}} + \frac{B}{{x - 1}} \cr
& 1 = A\left( {x - 1} \right) + B\left( {x + 5} \right) \cr
& {\text{let }}x = - 5 \cr
& 1 = A\left( { - 6} \right),{\text{ }}A = - \frac{1}{6} \cr
& {\text{let }}x = 1 \cr
& 1 = B\left( 6 \right),{\text{ }}B = \frac{1}{6} \cr
& \frac{1}{{\left( {x + 5} \right)\left( {x - 1} \right)}} = - \frac{1}{{6\left( {x + 5} \right)}} + \frac{1}{{6\left( {x - 1} \right)}} \cr
& ,{\text{ then}} \cr
& \int {\frac{{dx}}{{{x^2} + 4x - 5}}} = \int {\left( {\frac{1}{{6\left( {x - 1} \right)}} - \frac{1}{{6\left( {x + 5} \right)}}} \right)} dx \cr
& {\text{Integrating}} \cr
& = \frac{1}{6}\ln \left| {x - 1} \right| - \frac{1}{6}\ln \left| {x + 5} \right| + C \cr
& = \frac{1}{6}\ln \left| {\frac{{x - 1}}{{x + 5}}} \right| + C \cr
& {\text{Write in terms of }}\theta ,{\text{ replace }}x = \sin \theta \cr
& = \frac{1}{6}\ln \left| {\frac{{\sin \theta - 1}}{{\sin \theta + 5}}} \right| + C \cr
& = \frac{1}{6}\ln \left( {\frac{{1 - \sin \theta }}{{\sin \theta + 5}}} \right) + C \cr} $$
