Find $\int e^{x} \sin x \, dx$.

Let $I = \int e^{x} \sin x \, dx$. Using the integration by parts formula $\int u \, dv = uv - \int v \, du$,let $u = \sin x$ and $dv = e^{x} \, dx$. Then $du = \cos x \, dx$ and $v = e^{x}$.
$I = e^{x} \sin x - \int e^{x} \cos x \, dx$ ..........$(1)$
Now,evaluate $\int e^{x} \cos x \, dx$ using integration by parts again. Let $u = \cos x$ and $dv = e^{x} \, dx$. Then $du = -\sin x \, dx$ and $v = e^{x}$.
$\int e^{x} \cos x \, dx = e^{x} \cos x - \int e^{x} (-\sin x) \, dx = e^{x} \cos x + \int e^{x} \sin x \, dx$
Substituting this back into equation $(1)$:
$I = e^{x} \sin x - (e^{x} \cos x + I)$
$I = e^{x} \sin x - e^{x} \cos x - I$
$2I = e^{x} (\sin x - \cos x)$
$I = \frac{e^{x}}{2} (\sin x - \cos x) + C$