(e ^ x) * (sinx) ^ 2的不定积分

首先可以尝试使用分部积分法来求解这个不定积分。设$u = \sin x$，$dv = e^x(\sin x)^2dx$，则$du = \cos x dx$，$v = \int e^x(\sin x)^2dx$。于是有：

$$
\begin{aligned}
\int e^x(\sin x)^2dx &= \int u dv \\
&= uv - \int v du \\
&= e^x(\sin x)^2 - \int e^x \sin x \cos x dx \\
&= e^x(\sin x)^2 + \frac{1}{2}\int e^x \sin 2x dx \\
&= e^x(\sin x)^2 + \frac{1}{2}\int e^x \frac{\sin 2x}{2} d(2x) \\
&= e^x(\sin x)^2 + \frac{1}{4}\int e^x \sin 2x d(2x) \\
&= e^x(\sin x)^2 + \frac{1}{4}\int e^x \sin t dt \quad (t=2x) \\
&= e^x(\sin x)^2 - \frac{1}{4}e^x \cos x - \frac{1}{4}\int e^x \cos x dx \\
&= e^x(\sin x)^2 - \frac{1}{4}e^x \cos x - \frac{1}{4}\int e^x d(\sin x) \\
&= e^x(\sin x)^2 - \frac{1}{4}e^x \cos x - \frac{1}{4}e^x \sin x + C
\end{aligned}
$$

其中$C$为常数。因此，$\int e^x(\sin x)^2dx = e^x(\sin x)^2 - \frac{1}{4}e^x \cos x - \frac{1}{4}e^x \sin x + C$。