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 = (\left(\sin(\pi n) - \pi \cos(\pi n)\right) e^n)/(1+\pi^2) + C`