I was solving a physics problem and I run in to an integral that I could not solve. Could you give me some tips to solve this kind of integral?
For context I'm letting the original vectors at the end of the post, and the link to the physics problem.
\begin{align} V &= D \int_{\alpha_0}^{\alpha}\frac{\sec^2{\alpha} \ d\alpha}{\Vert\vec{r}-\vec{r_Q} \Vert} \tag{1} \\ \Vert\vec{r}-\vec{r_Q} \Vert &= \sqrt{r^2 + D^2 \sec^2{\alpha} - 2rD\ [\sin{\theta}\sin{\phi}\tan{\alpha} + \cos{\theta}]} \tag{2} \\ \alpha_0 &= \arctan(-\frac{L}{2D}) \qquad \alpha = \arctan(\frac{L}{2D}) \end{align}
$$ \vec{r} = r\sin\theta\cos\phi\ \hat{i}\ +\ r\sin\theta\sin\phi\ \hat{j}\ +\ r\cos\theta\ \hat{k} $$ $$ \vec{r_Q} = D\tan\alpha\ \hat{j}\ +\ D\ \hat{k} $$
This integral as stated can be evaluated using the substitution $u=\tan\alpha$, which brings it to the form
$\int_{-\delta}^{\delta}\frac{du}{\sqrt{au^2+bu+c}}=\frac{1}{\sqrt{a}}\sinh^{-1}\Bigg(\frac{u+\frac{b}{2a}}{\sqrt{\frac{4ac-b^2}{4a^2}}}\Bigg)\Bigg|^{\delta}_{-\delta}$
where $\delta=\frac{L}{2D}$, $a=D^2$ , $b=-2rD\sin\theta\sin\phi$ , $c=r^2+D^2-2rD\cos\theta$.
I took a look at the physics however and it seems that you have not considered the fact that the image of the charged rod through the conducting sphere is not going to be a rod itself, but a circle because every point of the rod is mapped a different distance away from the center of the sphere. Taking this into account, I don't know how useful this calculation is going to be for you but I submit it anyway.