# Why does exchange integral 1s2p not vanish for Helium excited state?

I am studying first-order perturbation theory applied to the first excited state of Helium. I am facing the following problem:

I understand that the Coulomb integral for all 2p functions will not be $$0$$, but for me, all the exchange integrals involving the 2p functions will be $$0$$. For example, look at this two integrals for the 1s2pz state:

1s function: $$1s=\sqrt{\frac{Z^3}{a_0^3 π}}e^{\frac{-Z}{a_0}r}$$ 2pz function: $$2p_z=\frac{1}4\sqrt{\frac{Z^5}{2a_0^5 π}}re^{\frac{-Z}{2a_0}r}cosθ$$

Coulomb integral: $$J_{p_z}=H_{(1s2pz;1s2pz)} =∫∫\frac{1s(1)2p_z(2)e^21s(1)2p_z (2)}{4πε_0r_{12}}dr_1 dr_2$$ $$J_{p_z}=\frac{Z^8 e^2}{128a_0^8 π^3ε_0} ∫_0^∞e^{{\frac{-2Z}{a_0}}r_1} r_1^2 dr_1 ∫_0^∞e^{{\frac{-2Z}{a_0}}r_2} r_2^4 dr_2 ∫_0^π\frac{senθ_1 dθ_1}{\sqrt{r_1^2+r_2^2-2r_1 r_2 cosθ_1}} ∫_0^{2π}dϕ_1 ∫_0^πcos^2 θ_2 senθ_2 dθ_2 ∫_0^{2π}dϕ_2$$ Exchange integral:

$$K_{p_z}=H_{(1s2pz;1s2pz)} =∫∫\frac{1s(1)2p_z(2)e^22p_z(1)1s(2)}{4πε_0r_{12}}dr_1 dr_2$$ $$K_{p_z}=\frac{Z^8 e^2}{128a_0^8 π^3ε_0} ∫_0^∞e^{{\frac{-3Z}{2a_0}}r_1} r_1^3 dr_1 ∫_0^∞e^{{\frac{-3Z}{2a_0}}r_2} r_2^3 dr_2 ∫_0^π\frac{cosθsenθ_1 dθ_1}{\sqrt{r_1^2+r_2^2-2r_1 r_2 cosθ_1}} ∫_0^{2π}dϕ_1 ∫_0^πcos θ_2 senθ_2 dθ_2 ∫_0^{2π}dϕ_2$$

In all cases, the electron 1 was placed along the z-axis for its $$θ$$ angle to be equal to the angle between the two electrons. So, the exchange integral (K) will be zero because the integral: $$∫_0^πcos θ_2 senθ_2 dθ_2=-\frac{(cos^2π-cos^20)}2=0$$ And the Coulomb Integral will not vanish because: $$∫_0^πcos^2 θ_2 senθ_2 dθ_2=-\frac{(cos^3π-cos^30)}3=\frac{2}3$$ But in all the materials I have read, both Coulomb and exchange integrals are not zero, but they do not show the calculations. I have searched a lot on the internet, and I still have not found where my mistake is.