Answer
\[\frac{\tan^2 x}{2}+\frac{\tan^4 x}{4}+C\]
Work Step by Step
Let \[I=\int\sin x\:\sec^5 xdx\]
\[I=\int\frac{\sin x}{\cos^5 x}dx\]
\[I=\int\tan x\:\sec^4 xdx\]
\[\left[\sec^2 x-\tan^2 x=1\right]\]
\[I=\int\tan x(1+\tan^2 x)\:\sec^2 xdx\]
Substitute $t=\tan x\;\; \Rightarrow dt=\sec^2 xdx$
\[I=\int t(1+t^2)dt\]
\[I=\int (t+t^3)dt\]
\[I=\frac{t^2}{2}+\frac{t^4}{4}+C\]
$C$ is constant of integration
\[I=\frac{\tan^2 x}{2}+\frac{\tan^4x}{4}+C\]
Hence,\[I=\frac{\tan^2 x}{2}+\frac{\tan^4x}{4}+C\].
