[tex]I=\int e^x(\sin(x)\cos(x))dx=\int e^x(\frac{1}{2}\sin(2x))dx=\frac{1}{2}\int e^x\sin(2x)dx[/tex]
[tex]\text{If }u=\sin(2x)\to du=2\cos(2x)dx~\text{and}~dv=e^xdx\to v=e^x:\\\\
\text{Using }\int u\,dv=uv-\int v\,du:\\\\
I=\frac{1}{2}(e^x\sin(2x)-\int e^x(2\cos(2x))dx)\\\\
2I=e^x\sin(2x)-2\underbrace{\int e^x\cos(2x)dx}_{I_2}[/tex]
Looking for [tex]I_2[/tex]:
[tex]\text{If}~u=\cos(2x)\to du=-2\sin(2x)dx~\text{and}~dv=e^xdx\to v=e^x:\\\\
I_2=e^x\cos(2x)-\int e^x(-2\sin(2x))dx\\\\ I_2=e^x\cos(2x)+2\int e^x(\sin(2x))dx\\\\ I_2=e^x\cos(2x)+2\int e^x(2\sin(x)\cos(x))dx\\\\ I_2=e^x\cos(2x)+4\int e^x(\sin(x)\cos(x))dx=e^x\cos(2x)+4I[/tex]
Replacing:
[tex]2I=e^x\sin(2x)-2I_2\iff\\\\2I=e^x\sin(2x)-2(e^x\cos(2x)+4I)\iff\\\\
2I=e^x\sin(2x)-2e^x\cos(2x)-8I\iff\\\\
10I=e^x\sin(2x)-2e^x\cos(2x)\iff\\\\
I=\dfrac{e^x}{10}(\sin(2x)-2\cos(2x))\\\\
\boxed{\int e^x(\sin(x)\cos(x))dx=\dfrac{e^x}{10}(\sin(2x)-2\cos(2x))+C}[/tex]
