Answer
$$ \frac{1}{2}[\sin x \sinh x-\cos x \cosh x]+C$$
Work Step by Step
Given $$ \int(\sin x)(\cosh x) d x $$
Let
\begin{align*}
u&=\sin x \ \ \ \ \ \ \ \ \ \ dv= \cosh x d x \\
du&=\cos xdx\ \ \ \ \ \ \ \ \ v=\sinh x
\end{align*}
Then
\begin{align*}
\int(\sin x)(\cosh x) d x&= \sin x\sinh x-\int\cos x \sinh xdx
\end{align*}
Let
\begin{align*}
u&=\cos x \ \ \ \ \ \ \ \ \ \ dv= \sinh x d x \\
du&=-\sin xdx\ \ \ \ \ \ \ \ \ v=\cosh x
\end{align*}
Then
\begin{aligned}
\int \sin x \cosh x \, d x &=\sin x \cdot \sinh x-\int \sinh x \cdot \cos x \, d x \\
&=\sin x \cdot \sinh x-\left[\cos x \cdot \cosh x-\int \cosh x \cdot(-\sin x) d x\right] \\
&=\sin x \cdot \sinh x-\left[\cos x \cdot \cosh x+\int \sin x \cosh x \, d x\right] \\
&=\sin x \sinh x-\cos x \cosh x-\int \sin x \cosh x \, d x
\end{aligned}
Hence
$$\int \sin x \cosh x \, d x=\frac{1}{2}[\sin x \sinh x-\cos x \cosh x]+C$$
