Answer
\[\frac{1}{a}\cosh au+C\]
Where $C$ is constant of integration
Work Step by Step
Let \[I=\int \sinh a u\,du\]
\[I=\int\left(\frac{e^{au}-e^{-au}}{2}\right)\,du\]
\[\Rightarrow I=\frac{1}{2}\int (e^{au}-e^{-au})du\]
\[\Rightarrow I=\frac{1}{2a} (e^{au}+e^{-au})+C\]
Where $C$ is constant of integration
\[I=\frac{1}{a}\left(\frac{e^{au}+e^{-au}}{2}\right)+C\]
\[I=\frac{1}{a}\cosh au+C\]
Hence , \[\int \sinh a u\,du=\frac{1}{2}\cosh au+C\]
