Answer
$$\ln \left| {\frac{{\sin x - 1}}{{\sin x + 4}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{5\cos x}}{{{{\sin }^2}x + 3\sin x - 4}}} dx \cr
& {\text{Let }}t = \sin x,{\text{ }}dt = \cos xdx \cr
& {\text{Substituting}} \cr
& \int {\frac{{5\cos x}}{{{{\sin }^2}x + 3\sin x - 4}}} dx = \int {\frac{5}{{{t^2} + 3t - 4}}} dt \cr
& {\text{Decompose into partial fractions}} \cr
& \frac{5}{{{t^2} + 3t - 4}} = \frac{5}{{\left( {t + 4} \right)\left( {t - 1} \right)}} = \frac{A}{{t + 4}} + \frac{B}{{t - 1}} \cr
& 5 = A\left( {t - 1} \right) + B\left( {t + 4} \right) \cr
& t = - 4 \to A = - 1 \cr
& t = 1 \to B = 1 \cr
& \frac{5}{{\left( {t + 4} \right)\left( {t - 1} \right)}} = \frac{{ - 1}}{{t + 4}} + \frac{1}{{t - 1}} \cr
& \int {\frac{5}{{{t^2} + 3t - 4}}} dt = \int {\left( {\frac{{ - 1}}{{t + 4}} + \frac{1}{{t - 1}}} \right)} dt \cr
& {\text{Integrate}} \cr
& {\text{ = }} - {\text{ln}}\left| {t + 4} \right| + \ln \left| {t - 1} \right| + C \cr
& {\text{Using logarithmic properties}} \cr
& = \ln \left| {\frac{{t - 1}}{{t + 4}}} \right| + C \cr
& {\text{Write in terms of }}x \cr
& = \ln \left| {\frac{{\sin x - 1}}{{\sin x + 4}}} \right| + C \cr} $$
