Answer
$\frac{1}{{15}}\sin y\left( {3{{\cos }^4}y + 4{{\cos }^2}y + 8} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {{{\cos }^5}y} dy \cr
& {\text{Using the entry 74 }}\left( {{\text{Integrals on Reference Pages 6}}--{\text{1}}0} \right) \cr
& {\text{Entry 74 }}\int {{{\cos }^n}udu} = \frac{1}{n}{\cos ^{n - 1}}u\sin u + \frac{{n - 1}}{n}\int {{{\cos }^{n - 2}}u} du \cr
& \cr
& {\text{Let }}u = y,{\text{ }}du = dy,{\text{ }}n = 5,{\text{ we obtain}} \cr
& \int {{{\cos }^5}y} dy = \frac{1}{5}{\cos ^{5 - 1}}y\sin y + \frac{{5 - 1}}{5}\int {{{\cos }^{5 - 2}}y} dy \cr
& \int {{{\cos }^5}y} dy = \frac{1}{5}{\cos ^4}y\sin y + \frac{4}{5}\int {{{\cos }^3}y} dy \cr
& \cr
& {\text{For }}\int {{{\cos }^3}y} dy{\text{ use the entry 68}} \cr
& {\text{Entry 68 }}\int {{{\cos }^3}udu} = \frac{1}{3}\left( {2 + {{\cos }^2}u} \right)\sin u + C \cr
& \int {{{\cos }^5}y} dy = \frac{1}{5}{\cos ^4}y\sin y + \frac{4}{5}\left[ {\frac{1}{3}\left( {2 + {{\cos }^2}y} \right)\sin y} \right] + C \cr
& \int {{{\cos }^5}y} dy = \frac{1}{5}{\cos ^4}y\sin y + \frac{4}{{15}}\left( {2 + {{\cos }^2}y} \right)\sin y + C \cr
& {\text{Factoring out }}\frac{1}{{15}}\sin y \cr
& \int {{{\cos }^5}y} dy = \frac{1}{{15}}\sin y\left[ {3{{\cos }^4}y + 4\left( {2 + {{\cos }^2}y} \right)} \right] + C \cr
& \int {{{\cos }^5}y} dy = \frac{1}{{15}}\sin y\left( {3{{\cos }^4}y + 4{{\cos }^2}y + 8} \right) + C \cr} $$
