Recall a few known results involving the Laplace transform. Given a function [tex]f(t)[/tex], if the transform exists, then denote it by [tex]F(s)[/tex]. We have
[tex]\mathcal L_s\left\{e^{ct}f(t)\right\}=\mathcal L_{s-c}\left\{f(t)\right\}=F(s-c)[/tex]
[tex]\mathcal L_s\left\{\displaystyle\int_0^t f(u)\,\mathrm du\right\}=\dfrac{F(s)}s[/tex]
[tex]\mathcal L_s\left\{f'(t)\right\}=sF(s)-f(0)[/tex]
Let's put all this together by taking the transform of both sides of the ODE:
[tex]y'(t)+2e^{-2t}\displaystyle\int_0^te^{2u}y(u)\,\mathrm du=e^{-t}\sin t[/tex]
[tex]\implies \bigg(sY(s)-y(0)\bigg)+2\mathcal L_s\left\{e^{-2t}\displaystyle\int_0^te^{2u}y(u)\,\mathrm du\right\}=\dfrac1{(s+1)^2+1}[/tex]
Here we use the third fact and immediately compute the transform of the right hand side (I'll leave that up to you).
Now we invoke the first listed fact:
[tex]\mathcal L_s\left\{e^{-2t}\displaystyle\int_0^te^{2u}y(u)\,\mathrm du\right\}=\mathcal L_{s+2}\left\{\displaystyle\int_0^te^{2u}y(u)\,\mathrm du\right\}[/tex]
Let [tex]g(u)=e^{2u}y(u)[/tex]. From the second fact, we get
[tex]\mathcal L_s\left\{\displaystyle\int_0^tg(u)\,\mathrm du\right\}=\dfrac{G(s)}s\implies\mathcal L_{s+2}\left\{\displaystyle\int_0^tg(u)\,\mathrm du\right\}=\dfrac{G(s+2)}{s+2}[/tex]
From the first fact, we get
[tex]G(s)=\mathcal L_s\left\{e^{2u}y(u)\right\}=Y(s-2)[/tex]
so we're left with
[tex]\dfrac{G(s+2)}{s+2}=\dfrac{Y((s+2)-2)}{s+2}=\dfrac{Y(s)}{s+2}[/tex]
To summarize, taking the Laplace transform of both sides of the ODE yields
[tex]sY(s)+\dfrac{2Y(s)}{s+2}=\dfrac1{(s+1)^2+1}[/tex]
Isolating [tex]Y(s)[/tex] gives
[tex]Y(s)=\dfrac1{\left(s+\frac2{s+2}\right)\left((s+1)^2+1\right)}[/tex]
[tex]Y(s)=\dfrac{s+2}{\left((s+1)^2+1\right)^2}[/tex]
All that's left is to take the inverse transform. I'll leave that to you as well. You should end up with something resembling
[tex]y(t)=\dfrac12(t\cos t-(t+1)\sin t)(\sinh t-\cosh t)[/tex]
