We're generating a sequence of functions [tex]Y_n[/tex] with
[tex]\begin{cases}Y_0(x)=y(0)\\\\Y_{n+1}(x)=y(0)+\displaystyle\int_{t=0}^{t=x}f(t,Y_n(t))\,\mathrm dt&\text{for }n\ge0\end{cases}[/tex]
where [tex]y'=f(x,y)[/tex], so that the sequence [tex]Y_n[/tex] converges to [tex]y[/tex] as [tex]n\to\infty[/tex].
[tex]n=0[/tex] :
[tex]Y_0(x)=1[/tex]
[tex]n=1[/tex] :
[tex]\displaystyle Y_1(x)=1+\int_0^x(t+\cos1)\,\mathrm dt=1+(\cos 1)x+\frac{x^2}2[/tex]
[tex]n=2[/tex] :
[tex]\displaystyle Y_2(x)=1+\int_0^x\left(t+\cos\left(1+(\cos 1)t+\frac{t^2}2\right)\right)\,\mathrm dt[/tex]
Unless you're familiar with Fresnel integrals, you won't be able to simplify this any further.
[tex]n=3[/tex] :
[tex]\displaystyle Y_3(x)=1+\int_0^x(t+\cos(Y_2(t)))\,\mathrm dt[/tex]
My computer takes a really long time to compute [tex]Y_3(x)[/tex], and even longer to plot it, so I've ultimately omitted [tex]Y_3[/tex] from the plot. (I wonder now if by "until to 3-th aproximation" it's intended that you only go up to [tex]n=2[/tex]...)
I've attached a plot of the approximations (dashed and colored) along with a more precisely computed numerical solution (black)
