# Integral Over A Hemishpere

1. Sep 17, 2007

### a_Vatar

Hi,
I'm trying to evaluate this integral over a hemisphere:
$$\int cos(\theta)^{a*cos^2(\phi) + b*sin^2(\phi)} dw$$
where dw - solid angle measure,$$\phi$$ is azimuthal angle and $$\theta$$
is polar angle.
Thus we have:
$$\int cos(\theta)^{a*cos^2(\phi) + b*sin^2(\phi)} dw$$ = $$\int \int cos(\theta)^{a*cos^2(\phi) + b*sin^2(\phi)} * sin(\theta) d\theta d\phi$$.
over hemishere.

Integral = $$\int^{2 * pi}_{0} \int^{pi/2}_{0}cos(\theta)^{a*cos^2(\phi) + b*sin^2(\phi)} * sin(\theta) d\theta d\phi$$

evaluate inner integral:
making substitution u = cos($$\theta$$),
$$\int^{pi/2}_{0}cos(\theta)^{a*cos^2(\phi) + b*sin^2(\phi)} * sin(\theta) d\theta = - \int^{pi/2}_{0}u^{a*cos^2(\phi) + b*sin^2(\phi)} du = \frac{1}{a*cos^2(\phi) + b*sin^2(\phi) + 1}$$

That's where I get stuck with integration wrt $$\phi$$
Apparently this integral has to evaluate to $$\frac{1}{\sqrt{(a+1)(b+1)}}$$

2. Sep 18, 2007

### AiRAVATA

Use the substitution

$$t=\tan \frac{\phi}{2},$$

and you end up with a rational integral.

3. Sep 20, 2007

### a_Vatar

$$\int \frac{1}{a*cos^2(\phi) + b*sin^2(\phi) + 1} d\phi$$

after letting $$t=\tan \frac{\phi}{2}$$ I get

$$\int \frac{1+t^2}{a(1-t^2)^2 + 4bt^2 + (1+t^2)^2}$$

this does not look to facinating, so I let $$u=t^2$$ and this simplifies it a bit further, but introduces $$\sqrt{u}$$ into denominator;

I tried pluging original $$\int \frac{1}{a*cos^2(\phi) + b*sin^2(\phi) + 1} d\phi$$
in the online integrator here (http://integrals.wolfram.com/index.jsp) and the result is reasonably simple, but when I plugged the "t subsitution integral", the result is way more complex! Btw, the u substitution apprarently yeilds result with complex number, where these come from I have no idea, cause the original integral is simple.

I don't know if Mathemathica actually can give you the intermediate steps of integration e.g. how it got the result, but if it can, could some one having access to it post integration process here plz?

4. Sep 20, 2007

### AiRAVATA

True! But if you want to solve it with real variable, now you can use partial fractions, i.e.

$$\frac{1+t^2}{a(1-t^2)^2 + 4bt^2 + (1+t^2)^2}=\frac{1+t^2}{(\sqrt{1+a}t^2+2\sqrt{a-b}t+\sqrt{1+a}) (\sqrt{1+a}t^2-2\sqrt{a-b}t+\sqrt{1+a})}$$

(assuming $a>b$), hence

$$\frac{1+t^2}{a(1-t^2)^2 + 4bt^2 + (1+t^2)^2}=\frac{1}{\sqrt{1+a}}\left( \frac{1}{\sqrt{1+a}t^2+2\sqrt{a-b}t+\sqrt{1+a}}+ \frac{1}{\sqrt{1+a}t^2-2\sqrt{a-b}t+\sqrt{1+a}}}\right).$$

Using complex variable is fair more simple, just do the substitution $2 \cos \phi=z+1/z,\,2i \sin \phi=z-1/z$ and use Cauchy's Theorem.

P.S. You are missing a factor of two multipliying the integral.

Last edited: Sep 20, 2007
5. Sep 21, 2007

### a_Vatar

How did you get this expression?

$$\frac{1+t^2}{(\sqrt{1+a}t^2+2\sqrt{a-b}t+\sqrt{1+a}) (\sqrt{1+a}t^2-2\sqrt{a-b}t+\sqrt{1+a})}$$

when I factor

$${a(1-t^2)^2 + 4bt^2 + (1+t^2)^2} = (a+1)t^4 -2(a-2b-1)t^2 + (a+1)$$
then say,

$$u=t^2$$ and
$$u1,2 = \frac{(a-2b-1)\pm 2\sqrt{(1+b)(b-1)}}{a+1}$$,
thus
$$(a+1)t^4 -2(a-2b-1)t^2 + (a+1) = (t^2 - u1)(t^2 - u2)$$

6. Sep 24, 2007

### AiRAVATA

Simply write the expression

$$(a+1)t^4 -2(a-2b-1)t^2 + (a+1)=(\sqrt{a+1}t^2+\alpha t+\sqrt{a+1})(\sqrt{a+1}t^2+\beta t +\sqrt{a+1})$$

and compute $\alpha,\,\beta$.

7. Sep 24, 2007

### a_Vatar

but where do this coefficient come from:
$$\sqrt{a+1}$$
Is there a general formula to expend a quartic? Or am I missing something obvious
Thanks.

8. Sep 24, 2007

### AiRAVATA

Well, I started trying to express the quartic as a product of two second order polinomials, ending up with five equations for six coefficients, wich meant there is no unique expression, so the logical approach was to use the symmetry of the highest and lowest term.