How to calculate this Integral which function is defined by parts

Question

Let $k:[0,1]\times [0,1] \to \mathbb{R}$ defined as: $k(x,y) = \left\{ \begin{array}{ll} y(1-x) & \mbox{if } 0\leq y\leq x \le 1,\\ x(1-y) & \mbox{if } 0\leq x\leq y \le 1. \end{array} \right.$

If we denote $u_n(x):=\sin(n\pi x), \forall n\in \mathbb{N}$. Prove that: $\int_0^1k(x,y)u_n(y) dy=\dfrac{\sin(n\pi x)}{n^2\pi^2}.$

My attempt was integrating by parts, and see that $k(x,0)=k(x,1)=0$ and $k'(x,y)=f(y)$ i.e the derivate of $f$ doesn't deppends of the value of $x$ at all, but is defined by parts.

But what I got using the previous facts was that:

$\int_0^1k(x,y)u_n(y) dy= \dfrac{-x\sin(n\pi)}{n^2\pi^2}=0, $ since $n\in \mathbb{N}.$

So... What I'm doing wrong, maybe Integrating By Parts in a function defined my parts, or maybe I may use theorems about compactness or convergent sub-sequences because $(u_n)$ is bounded and k is uniformly continuous. Any idea will be appreciated.

Answer 1

Proceed as follows \begin{align}\int_0^1k(x,y)u_n(y) dy =& \int_0^1k(x,y)\sin(n\pi y)dy\\ =& \>(1-x)\int_0^x y \sin(n\pi y)dy + x\int_x^1 (1-y) \sin(n\pi y)dy\\ \overset{ibp}=&\> (1-x)\left( -\frac{x\cos n\pi x}{n\pi}+\int_0^x \frac{\cos n\pi y}{n\pi}dy\right)\\ &\hspace{5mm}+x\left( \frac{(1-x)\cos n\pi x}{n\pi}-\int_x^1\frac{\cos n\pi y}{n\pi}dy\right) \\=&\>(1-x)\int_0^x \frac{\cos n\pi y}{n\pi}dy -x\int_x^1 \frac{\cos n\pi y}{n\pi}dy \\ = &\>\int_0^x \frac{\cos n\pi y}{n\pi}dy -x\int_0^1 \frac{\cos n\pi y}{n\pi}dy \\ =&\>\dfrac{\sin(n\pi x)}{n^2\pi^2} \end{align}