Evaluate $\int e^x \sin^2 x \mathrm{d}x$ Is the following evaluation of correct?
\begin{align*} \int e^x \sin^2 x \mathrm{d}x &= e^x \sin^2 x -2\int e^x \sin x \cos x \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (\cos^2 x - \sin^2x) \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (1 - 2\sin^2x) \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int -2 e^x  \sin^2x \mathrm{d}x + 2 e^x \\ &= e^x \sin^2 x -2e^x \sin x \cos x  + 2 e^x -4 \int e^x  \sin^2x \mathrm{d}x \end{align*}
First two steps use integration by parts. In the first step we differentiate $\sin^2 x$. In the second step we differentiate  $\sin x \cos x$.
Using this, we reach $$5\int e^x \sin^2 x \mathrm{d}x = e^x \sin^2 x -2e^x \sin x \cos x  + 2 e^x$$
$$\int e^x \sin^2 x \mathrm{d}x = \frac{e^x \sin^2 x -2e^x \sin x \cos x  + 2 e^x}{5}+C$$
I can't reach the form that most integral calculators give, which has terms $\cos(2x)$ and $\sin(2x)$ by just using trig identities, so I wonder whether the result is correct. I would also be interested in a method that immediately gives the form $$-\frac{e^x[2 \sin(2x)+ \cos(2x)-5]}{10}+C$$

$$\int \left(e^x\sin^2(x)\right)\text{d}x =$$
$$\int \left(e^x\left(\frac{1}{2}(1-\cos(2x))\right)\right)\text{d}x =$$
$$\frac{1}{2}\int \left(e^x-e^x\cos(2x)\right)\text{d}x =$$
$$\frac{1}{2} \left(\int \left(e^x\right) \text{d}x-\int \left(e^x\cos(2x)\right) \text{d}x\right) =$$
$$\frac{1}{2} \left(\int e^x \text{d}x-\int e^x\cos(2x) \text{d}x\right) =$$
$$\frac{1}{2} \left(e^x-\int e^x\cos(2x) \text{d}x\right) =$$

For the integrand $e^x\cos(2x)$, use the formula:
$$\int\exp(\alpha x)\cos(\beta x)\text{d}x=\frac{\exp(\alpha x)(\alpha \cos(\beta x))+\beta\sin(\beta x)}{\alpha^2+\beta^2}$$

$$\frac{1}{2} \left(e^x-\frac{e^x(2\sin(2x)+\cos(2x))}{5}\right) + C =$$
$$-\frac{e^x(2\sin(2x)+\cos(2x)-5)}{10} + C $$
