Answer
$-\displaystyle \frac{1}{2}\cosh(1-2x)+C$
Work Step by Step
Use Th.5.18, Derivatives and Integrals of Hyperbolic Functions
$\displaystyle \int\sinh udu=\cosh u+C$
----
$\displaystyle \int\sinh(1-2x)dx=\left[\begin{array}{lll}
u=1-2x & & \\
du=-2dx. & \Rightarrow & -\frac{1}{2}du=dx
\end{array}\right]$
$=-\displaystyle \frac{1}{2}\int\sinh\underbrace{(1-2x)}_{u}\underbrace{(-2)dx}_{du}$
$=-\displaystyle \frac{1}{2}\cosh(1-2x)+C$
