I am stuck at simplifying the following integral: $$ I = \int_{-\pi/2}^{+\pi/2} d\phi \left[ 1-M(\phi) \theta^*(\phi) \right] $$ where $ M= p \cos \phi + q \sin \phi $. Here $p,q$ are fixed.
The $\theta^\star$ is defined through $\cos \theta^*= \frac{M}{\sqrt{1+M^2}} $
Can anybody help me to clarify things? I have reason to believe it should take a much nicer form.
Thank you in advance.
UPDATE The integral writes $ I= \pi- I_1$ where $I_1 = \int_{-\pi/2}^{+\pi/2} d\phi M(\phi) \theta^*(\phi) $
Using integration by parts as suggested, we obtain \begin{eqnarray} u =\theta^*(\phi) =\arccos G &\Rightarrow& du = \frac{-dG}{\sqrt{1-G^2}}= -\sqrt{1+M^2}dG= \frac{-A}{(1+M^2)}d\phi \\ dv=M(\phi) d\phi &\Rightarrow& v = -A \end{eqnarray}
Let $G=\frac{M}{\sqrt{1+M^2}}$ thus $dG =\frac{1}{(1+M^2)^{3/2}}dM =\frac{A}{(1+M^2)^{3/2}}d\phi $
where $ A(\phi)= M'(\phi)= q\cos \phi-p\sin \phi$.
Note that $A^2+M^2=p^2+q^2$.
Introduce $\alpha=\sqrt{p^2+q^2+1}$
Thus the integral writes \begin{eqnarray} I_1 &=& \left[ -\theta^* A \right]_{-\pi/2}^{+\pi/2} - \int_{-\pi/2}^{+\pi/2} d\phi \frac{A^2}{1+M^2} \\ &=& p\pi - \int_{-\pi/2}^{+\pi/2} d\phi \frac{A^2}{\alpha^2-A^2} \\ &=& p\pi +\pi - \int_{-\pi/2}^{+\pi/2} d\phi \frac{\alpha^2}{\alpha^2-A^2} \end{eqnarray}
I have reasons to think that the right integral simplifies into $$ \int_{-\pi/2}^{+\pi/2} d\phi \frac{\alpha^2}{\alpha^2-A^2} = \alpha \pi $$ but I could not manage to prove it. Any help would be highly appreciated.
