Finding the electric field at a point from half of a uniformly charged rod

The image below is a rod of length $L$ that is uniformly charged to total charge $Q$ at a point along the perpendicular bisector of the rod above distance $h$.

I know the formula for finding the electric field at a point charge is $E=kq/r^2$. I derived $q$ to get $dq=lambda dx$. To find $r$ I get $sqrt{x^2+h^2}$ and I know the angle is $cos(h/r)$. By symmetry the $E_x$ forces cancel each other out.

For the charge $Q$ at the point I can find the electric field by using these equations and deriving the electric field:

$$|dE_{yr}|=frac{klambda dx}{r^2}cos(h/r)=frac{klambda dx}{h^2+x^2}cdot frac{h}{(h^2+x^2)^{1/2}}=int_{0}^{2klambda h}frac{1}{(h^2+x^2)^{3/2}}dx=bigg[2klambda h[frac{x}{h^2cdot(h^2+x^2)^{1/2}}]bigg]_0^{L/2}$$

Finally equals:

$$|E|=frac{klambda L}{hcdot(h^2+(L/2)^2)^{3/2}}$$

However, now I am supposed to find the electric field due to half of this uniformly charged rod for total charge $Q$ at point $P$ shown in the picture below:

The $E_x$ forces do not cancel each other out. So far I know that it has both an $x$ and $y$ component.

I have attempted to find the $y$-component of the field:

$$|dE_yr|=frac{klambda dx}{r^2}cos(d/r)=frac{klambda dx}{d^2+x^2}cdotfrac{d}{(d^2+x^2)^{1/2}}=int_0^{klambda h}frac{1}{(d^2+x^2)^{3/2}}dx$$

and also attempted to find the $x$-component of the field:

$$|dE_xr|=frac{klambda dx}{r^2}sin(d/r)=frac{klambda dx}{d^2+x^2}cdotfrac{d}{(d^2+x^2)^{1/2}}=int_0^{klambda h}frac{1}{(d^2+x^2)^{3/2}}dx$$

did I set up the integrals right?