1. We want to evaluate the integral $$I = \int e^x \sin x \, dx.$$\n\n2. Use integration by parts. Let $$u = \sin x$$ and $$dv = e^x dx$$ so that $$du = \cos x dx$$ and $$v = e^x.$$\n\n3. Applying integration by parts formula: $$I = uv - \int v \, du = e^x \sin x - \int e^x \cos x \, dx.$$\n\n4. Let $$J = \int e^x \cos x \, dx.$$ We compute $$J$$ similarly by parts: let $$u = \cos x$$ and $$dv = e^x dx,$$ so $$du = -\sin x dx$$ and $$v = e^x.$$\n\n5. Then $$J = e^x \cos x - \int e^x (-\sin x) \, dx = e^x \cos x + I.$$\n\n6. Substitute $$J$$ back into the expression for $$I$$: $$I = e^x \sin x - J = e^x \sin x - (e^x \cos x + I) = e^x \sin x - e^x \cos x - I.$$\n\n7. Rearranging terms: $$I + I = 2I = e^x (\sin x - \cos x).$$\n\n8. Solve for $$I$$: $$I = \frac{e^x}{2} (\sin x - \cos x) + C,$$ where $$C$$ is the constant of integration.\n\nFinal answer: $$\int e^x \sin x \, dx = \frac{e^x}{2} (\sin x - \cos x) + C.$$