# Help with a simple cos substitution

1. Apr 6, 2013

### ck99

1. The problem statement, all variables and given/known data

Hi folks, I am sure this is very simple but there are not enough steps given in this calculation for my simple brain to get from the beginning to the end!

σ = ∫ (dσ/dΩ) = ∫ r2sin2θ (no integral limits given)

σ = 2∏r2 ∫ (1 - u2) du (integral from -1 to 1)

σ = 8∏r2 / 3

2. Relevant equations

u = cos θ

3. The attempt at a solution

I used Ω = sin θ dθ d∅ and first integrated ∅ from 0 to 2∏ to get

σ = ∫ dσ = r2 ∫ sin2θ dΩ

σ = 2∏ r2 ∫ sin2θ sin θ dθ

Use sin2θ = 1 - cos2θ to get

σ = 2∏ r2 ∫(1 - cos2θ) sin θ dθ

Let u = cos θ so du/dθ = - sin θ and dθ = -arcsin θ to get

σ = 2∏ r2 ∫(1 - u2) sin θ -arcsin θ du

I think sin and arcsin cancel to give

σ = 2∏ r2 -∫(1 - u2) du

σ = 2∏ r2 2u

From the answer that was given I have the integral limits running from -1 to 1 so the final term becomes [2 - (-2)] = 4 which gives

σ = 8∏ r2

Where am I going wrong please?

2. Apr 6, 2013

### ck99

Oh. I differentiated u instead of integrating. Some days there is not enough coffee in the world . . .

3. Apr 6, 2013

### I like Serena

Hi ck99!

You seem to mix up differentials a bit here and elsewhere.

I'm going to assume you actually intended:

σ = ∫ dσ = ∫ r2sin2θ dΩ

Here's another mix-up.
Differentials always need to balance.

This is not the case in dθ = -arcsin θ, but then, you don't need it.

Let's stick to: u = cos θ so du = - sin θ dθ.

That should be:

σ = 2∏ r2 ∫(1 - u2)(-du)