Here is the related problem:
Let $\{f_{n}:[0,1]\to\mathbb{R}\}$ be a sequence of functions defined by $$f_{n}(x)=\int_{0}^{x}\left\{1+(x-y)^{n}\cdot\sin^{n}(xy)\right\}\,dy$$ for each $n\in\mathbb{N}$. Prove that $\{f_{n}\}$ converges uniformly to some function $f:[0,1]\to\mathbb{R}$ on $[0,1]$.
My first attempt was as follows:
Since $f_{n}(0)=0$ and $f_{n}'(x)=1$ on $[0,1]$ by the F.T.C. for each $n\in\mathbb{N}$, $f_{n}'(x)$ converges uniformly to the constant $1$ on $[0,1]$.
Hence, $\{f_{n}\}$ converges uniformly to the function $f(x)=x$ on $[0,1]$ by the well-known theorem(for differentiablility with uniform convergence).
But, I'm not sure whether it could apply the F.T.C. to $f_{n}$'s, because the integrand have mixed variables $x$ and $y$.
(Here, "F.T.C." means the Fundamental Theorem of Calculus)
Can anyone criticize my attempt if it is a completely wrong argument, or notice me if there is a different version of the fundamental theorem of calculus for this problem. Thank you!
Claim: $f_n\to f$ uniformly on $[0,1]$, where $f(x)=x$.
For each $x\in [0,1]$, we have $$|f_n(x)-f(x)|=\left|\int_0^x[1+(x-y)^n\sin^n(xy)]\,dy-\int_0^xdy\right|\\\le\int_0^x|x-y|^n|\sin(xy)|^ndy\\\le\int_0^x|\sin(xy)|^ndy$$ since for each $x,y\in[0,1]$ we must have $|x-y|\le 1$. Now, for each $x,y\in [0,1]$, it is clear that $xy\in [0,1]$, and so $\sin(xy)\in [0,\sin 1]$. Thus, $$\int_0^x|\sin(xy)|^ndy\le x(\sin 1)^n\le(\sin 1)^n.$$ Since $1\in (0,\pi/2)$, $\sin 1\in (0,1)$. So for an arbitrary $\epsilon>0$, there exists some $N\in\mathbb N$ such that for all $n\ge N$ we have $(\sin 1)^n<\epsilon$. ...