How to integrate ∫ e^x sin(x) dx using integration by parts?

This is a classic example of a **cyclic integration by parts** problem. The trick is to treat the integral as an unknown variable and solve algebraically. Let $I = \displaystyle\int e^x \sin(x) \, dx$. Apply integration by parts with: \begin{align*} u &= \sin(x), & dv &= e^x \, dx \\ du &= \cos(x) \, dx, & v &= e^x \end{align*} Then: $$I = e^x \sin(x) - \int e^x \cos(x) \, dx$$ Now apply integration by parts again on $J = \displaystyle\int e^x \cos(x) \, dx$: \begin{align*} u &= \cos(x), & dv &= e^x \, dx \\ du &= -\sin(x) \, dx, & v &= e^x \end{align*} So: $$J = e^x \cos(x) - \int e^x (-\sin(x)) \, dx = e^x \cos(x) + I$$ Substituting back: $$I = e^x \sin(x) - \left(e^x \cos(x) + I\right)$$ $$I = e^x \sin(x) - e^x \cos(x) - I$$ $$2I = e^x (\sin(x) - \cos(x))$$ Therefore: $$\boxed{\int e^x \sin(x) \, dx = \frac{e^x}{2}(\sin(x) - \cos(x)) + C}$$