<s>[INST] Find reduction formula using integration by parts 
Let $\displaystyle I(n,\alpha)=\int_0^\pi \sin(n\theta)e^{\alpha \theta} d\theta$. Show that $\displaystyle I(n,\alpha)=\frac{n}{n^2+\alpha^2}(1-(-1)^ne^{\alpha \pi})$.

This is done by parts, but I can't get the expression to equal the RHS.
 [/INST] Note: I adapted this from Arturo's answer at: Integration by parts: $\int e^{ax}\cos(bx)\,dx$
Also, please forgive me for not using your variable names as I found them confusing (easy to modify the entire result by just changing them to your names).
Note: we are going to save the integration limits to the end for ease.
So, doing IBP twice yileds:
$$\int e^{ax}\sin(bx)dx = \frac{1}{a}e^{ax}\sin(bx) - \frac{b}{a}\left(\frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a}\int e^{ax}\cos(bx)\,dx\right).$$
Multiplying out you get
$$\int e^{ax}\sin(bx)\,dx = \frac{1}{a}e^{ax}\sin(bx) - \frac{b}{a^2}e^{ax}\cos(bx) - \frac{b^2}{a^2}\int e^{ax}\cos(bx)\,dx.$$
At this point, you should move that last integral on the right hand side to the left hand side and "leave oout" the constant of integration on the right (because we'll use limits at the end as noted earlier).
Moving the last integral to the left hand side, you get
$$\left(1 + \frac{b^2}{a^2}\right)\int e^{ax}\sin(bx)\,dx = \frac{1}{a}e^{ax}\sin(bx) - \frac{b}{a^2}e^{ax}\cos(bx).$$
Now, $\displaystyle 1 + \frac{b^2}{a^2} = \frac{a^2+b^2}{a^2}$, and we want to clear this from the left side, so multiply both sides by the reciprocal, that is by, $\displaystyle \frac{a^2}{a^2+b^2}$.
If you do that, from
$$\frac{a^2+b^2}{a^2}\int e^{ax}\sin(bx)\,dx = \frac{1}{a}e^{ax}\sin(bx) - \frac{b}{a^2}e^{ax}\cos(bx),$$
multiplying both sides by $\frac{a^2}{a^2+b^2}$, we get:
$$\tag 1 \int e^{ax}\sin(bx)\,dx = \frac{a}{a^2+b^2}e^{ax}\sin(bx) - \frac{b}{a^2+b^2}e^{ax}\cos(bx).$$
Now, we just need to deal with the RHS of $(1)$ with the limits from $\Pi$ to $0$.
So we have:
$\displaystyle \frac{a}{a^2+b^2}e^{ax}\sin(bx) - \frac{b}{a^2+b^2}e^{ax}\cos(bx)$ evaluated at $\pi$, yields: $\displaystyle \frac{a}{a^2+b^2} e^{a \pi}(a \sin b \pi - b \cos b \pi)$
Of course, $\sin b \pi= 0$ for all $b$ (odd or even) and $\cos b \pi = \pm 1$ (depending on parity of b, that is $-1$ if $b$ is odd and is $+1$ if b is even, so we will account for this by using $(-1)^b$).
$\displaystyle \frac{a}{a^2+b^2}e^{ax}\sin(bx) - \frac{b}{a^2+b^2}e^{ax}\cos(bx)$ evaluated at $0$, yields: $\frac{a}{a^2+b^2} (0 - b).$
Subtracting these expressions (because we are doing over limits $\pi$ and $0$), yields:
$\displaystyle \frac{a}{a^2+b^2}e^{ax}\sin(bx) - \frac{b}{a^2+b^2}e^{ax}\cos(bx)|{_0^\pi} = \frac{b}{a^2+b^2}\left(1 - (-1)^b e^{a \pi}\right)$.
The desired result.
Regards
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