Answer
$$ - \frac{{{4^{\cot x}}}}{{\ln 4}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{4^{\cot x}}}}{{{{\sin }^2}x}}dx} \cr
& {\text{trigonometric identity }}\csc x = \frac{1}{{\sin x}} \cr
& \int {{4^{\cot x}}{{\csc }^2}xdx} \cr
& {\text{substitute }}u = \cot x,{\text{ }}du = - {\csc ^2}xdx \cr
& \int {{4^{\cot x}}{{\csc }^2}xdx} = - \int {{4^u}du} \cr
& {\text{find the antiderivative}} \cr
& {\text{by the formula }}\int {{a^u}} du = \frac{{{a^u}}}{{\ln a}} + C \cr
& {\text{letting }}a = 4 \cr
& = - \frac{{{4^u}}}{{\ln 4}} + C \cr
& = - \frac{{{4^{\cot x}}}}{{\ln 4}} + C \cr} $$
