Answer
$$ - \frac{1}{4}\ln \left| {\sin x} \right| + \frac{1}{8}\ln \left| {\left( {\sin x + 2} \right)\left( {\sin x - 2} \right)} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\cos \theta }}{{\left( {{{\sin }^3}\theta - 4\sin \theta } \right)}}} d\theta \cr
& {\text{set }}u = \sin \theta ,{\text{ }}du = \cos \theta d\theta \cr
& \int {\frac{{\cos \theta }}{{\left( {{{\sin }^3}\theta - 4\sin \theta } \right)}}} d\theta = \int {\frac{{du}}{{{u^3} - 4u}}} \cr
& {\text{partial fractions}} \cr
& \frac{1}{{{u^3} - 4u}} = \frac{1}{{u\left( {{u^2} - 4} \right)}} = \frac{1}{{u\left( {u + 2} \right)\left( {u - 2} \right)}} \cr
& \frac{1}{{u\left( {u + 2} \right)\left( {u - 2} \right)}} = \frac{A}{u} + \frac{B}{{u + 2}} + \frac{C}{{u - 2}} \cr
& 1 = A\left( {u + 2} \right)\left( {u - 2} \right) + Bu\left( {u - 2} \right) + Cu\left( {u + 2} \right) \cr
& u = 0 \to 1 = A\left( { - 4} \right) \cr
& A = - \frac{1}{4} \cr
& u = - 2 \to 1 = B\left( { - 2} \right)\left( { - 2 - 2} \right) \cr
& B = \frac{1}{8} \cr
& u = 2 \to 1 = C\left( 2 \right)\left( {2 + 2} \right) \cr
& C = \frac{1}{8} \cr
& \frac{A}{u} + \frac{B}{{u + 2}} + \frac{C}{{u - 2}} = - \frac{1}{{4u}} + \frac{1}{{8\left( {u + 2} \right)}} + \frac{1}{{8\left( {u - 2} \right)}} \cr
& = \int {\frac{{du}}{{{u^3} - 4u}}} = \int {\left( { - \frac{1}{{4u}} + \frac{1}{{8\left( {u + 2} \right)}} + \frac{1}{{8\left( {u - 2} \right)}}} \right)du} \cr
& {\text{find the antiderivative}} \cr
& = - \frac{1}{4}\ln \left| u \right| + \frac{1}{8}\ln \left| {u + 2} \right| + \frac{1}{8}\ln \left| {u - 2} \right| + C \cr
& = - \frac{1}{4}\ln \left| u \right| + \frac{1}{8}\ln \left| {\left( {u + 2} \right)\left( {u - 2} \right)} \right| + C \cr
& {\text{with }}u = \sin x \cr
& = - \frac{1}{4}\ln \left| {\sin x} \right| + \frac{1}{8}\ln \left| {\left( {\sin x + 2} \right)\left( {\sin x - 2} \right)} \right| + C \cr} $$
