Question

Why is the answer to this integral's denominator have 1+pi^2

Answer 1

It comes from integrating by parts twice. Let

$I = \displaystyle \int e^n \sin(\pi n) \, dn$

Recall the IBP formula,

$\displaystyle \int u \, dv = uv - \int v \, du$

Let

$u = \sin(\pi n) \implies du = \pi \cos(\pi n) \, dn$

$dv = e^n \, dn \implies v = e^n$

Then

$\displaystyle I = e^n \sin(\pi n) - \pi \int e^n \cos(\pi n) \, dn$

Apply IBP once more, with

$u = \cos(\pi n) \implies du = -\pi \sin(\pi n) \, dn$

$dv = e^n \, dn \implies v = e^n$

Notice that the ∫ v du term contains the original integral, so that

$\displaystyle I = e^n \sin(\pi n) - \pi \left(e^n \cos(\pi n) + \pi \int e^n \sin(\pi n) \, dn\right)$

$\displaystyle I = \left(\sin(\pi n) - \pi \cos(\pi n)\right) e^n - \pi^2 I$

$\displaystyle (1 + \pi^2) I = \left(\sin(\pi n) - \pi \cos(\pi n)\right) e^n$

$\implies \displaystyle I = \frac{\left(\sin(\pi n) - \pi \cos(\pi n)\right) e^n}{1+\pi^2} + C$