1. The problem is to find the integral of the function $e^x \sin x$, i.e., compute $$\int e^x \sin x \, dx.$$\n\n2. We use the method of integration by parts, which states: $$\int u \, dv = uv - \int v \, du.$$\n\n3. Let $u = \sin x$ and $dv = e^x dx$. Then, $du = \cos x \, dx$ and $v = e^x$.\n\n4. Applying integration by parts: $$\int e^x \sin x \, dx = e^x \sin x - \int e^x \cos x \, dx.$$\n\n5. Now, we need to evaluate $$\int e^x \cos x \, dx.$$ We apply integration by parts again with $u = \cos x$, $dv = e^x dx$, so $du = -\sin x \, dx$, $v = e^x$.\n\n6. This gives: $$\int e^x \cos x \, dx = e^x \cos x - \int e^x (-\sin x) \, dx = e^x \cos x + \int e^x \sin x \, dx.$$\n\n7. Substitute this back into the original integral expression: $$\int e^x \sin x \, dx = e^x \sin x - \left(e^x \cos x + \int e^x \sin x \, dx\right) = e^x \sin x - e^x \cos x - \int e^x \sin x \, dx.$$\n\n8. Let $I = \int e^x \sin x \, dx$. Then, from above: $$I = e^x \sin x - e^x \cos x - I.$$\n\n9. Add $I$ to both sides: $$2I = e^x (\sin x - \cos x).$$\n\n10. Divide both sides by 2: $$I = \frac{e^x (\sin x - \cos x)}{2} + C,$$ where $C$ is the constant of integration.\n\nFinal answer: $$\int e^x \sin x \, dx = \frac{e^x (\sin x - \cos x)}{2} + C.$$