Wave function in a hydrogen atom : normalization

[Another]

Homework Statement

Determined wave function in a hydrogen atom.
$Ψ(r,θ,Φ) = A(x+iy)e^{ \frac{-r}{2a_0}}$ << find A by normalization

Answer of a question in my book is $A = -\frac{1}{a_0 \sqrt{8 \pi}} (\frac{1}{2a_0})^{3/2}$

Homework Equations

$\int Ψ^*(r,θ,Φ)Ψ(r,θ,Φ) d^3r = \int \int \int Ψ^*(r,θ,Φ)Ψ(r,θ,Φ) r^2 \sin (θ) drdθdΦ =1$
$x = r \sinθ \cos Φ$
$y = r \sinθ \sin Φ$
$z = r \cosθ$

The Attempt at a Solution

r [0→∞] ,θ[-$\pi$→$\pi$] ,Φ[0→$2 \pi$][/B]
$\int Ψ^*(r,θ,Φ)Ψ(r,θ,Φ) d^3r = \int \int \int e^{ \frac{-r}{a_0}} r^2(\sin^2θ \cos^2Φ + \sin^2θ \sin^2Φ) r^2 \sin θ drdθdΦ$
$\int Ψ^*(r,θ,Φ)Ψ(r,θ,Φ) d^3r = \int \int \int e^{ \frac{-r}{a_0}} r^4\sin^3θ drdθdΦ$
$\int Ψ^*(r,θ,Φ)Ψ(r,θ,Φ) d^3r = 2\pi \int e^{ \frac{-r}{a_0}} r^4 dr \int \sin^3θ dθ$
$\int Ψ^*(r,θ,Φ)Ψ(r,θ,Φ) d^3r = 2\pi a_0^5 4! \int \sin^3θ dθ$

$\int \sin^3θ dθ = 0$ ; θ[-$\pi$→$\pi$] because $sin^3θ$ is odd function

$\int Ψ^*(r,θ,Φ)Ψ(r,θ,Φ) d^3r = 0$ My answer is wrong , please help

 

[stevendaryl]

$\int \sin^3θ dθ = 0$ ; θ[-$\pi$→$\pi$] because $sin^3θ$ is odd function

$\int Ψ^*(r,θ,Φ)Ψ(r,θ,Φ) d^3r = 0$ My answer is wrong , please help

The angle $\theta$ doesn't go from 0 to $2\pi$, it only goes from 0 to $\pi$. To integrate $sin^3(\theta)$, change variables by letting $u = cos(\theta)$.

 

[Another]

The angle $\theta$ doesn't go from 0 to $2\pi$, it only goes from 0 to $\pi$. To integrate $sin^3(\theta)$, change variables by letting $u = cos(\theta)$.

$\int sin^3(\theta) d\theta$ from 0 to $\pi$ =$\frac{4}{3}$

thank you very much. This is my mistake.