Answer
$\frac{\sec ^4x}{4} + C$
Work Step by Step
$$\eqalign{
&\text{Let } I= \int {\sin x{{\sec }^5}x} dx \cr
& {\text{Use the reciprocal identity }}\sec x = \frac{1}{{\cos x}} \cr
& \int {\sin x{{\sec }^5}x} dx = \int {\sin x\left( {\frac{1}{{{{\cos }^5}x}}} \right)} dx \cr
& {\text{Integrate by the substitution method}} \cr
& {\text{Let }}u = \cos x,{\text{ }}du = - \sin xdx,{\text{ }} - du = \sin xdx \cr
& {\text{Substituting}} \cr
& \int {\sin x\left( {\frac{1}{{{{\cos }^5}x}}} \right)} dx = \int {\left( {\frac{1}{{{u^5}}}} \right)} \left( { - 1} \right)du \cr
& = - \int {\frac{1}{{{u^5}}}} du \cr
& = - \int {{u^{ - 5}}} du \cr
& {\text{Use the power rule }}\int {{u^n}} du = \frac{{{u^{n + 1}}}}{{n + 1}} + C \cr
& = - \frac{{{u^{ - 4}}}}{{ - 4}} + C \cr
& = \frac{1}{{{4u^4}}} + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}\cos x{\text{ for }}u \cr
& = \frac{1}{{{{4\cos }^4}x}} + C \cr
& = \frac{{\sec ^4}x}{4} + C \cr} $$
