Can you help me evaluating this integral? (no symbolic software please).
\begin{equation} \int \frac{C \; d \theta}{ \sin \theta \sqrt{\sin^2 \theta - C }} \end{equation}
It can be reduced to an algebraic form \begin{equation} \int \frac{dx}{ x \sqrt{ a + b x^2 + c x^4}} \end{equation} How about evaluating the second integral?
Thanks.
Actually I like very much the substitutions suggested by @juanheron. However he rushed a bit at the end and introduced errors on the result. In addition he left the work on half.
Let me start where Juan introduced the error:
Let us make the substitution $v^2 = k u^2 -1 $, so $ v dv = k \, u \, du$, and $u^2 -1 = (v^2 +1)/k -1 = (v^2 +1 - k)/k$, so \begin{equation} I = C \int \frac{v dv/k}{(v^2+1-k)/k) v} = C \int \frac{dv}{(v^2+1-k)} = C \int \frac{dv}{(v^2+C)} \end{equation} That is, \begin{equation} I = \int \frac{dv}{(v/\sqrt{C})^2 + 1}. \end{equation} A final substitution of $v/\sqrt{C}=z$, $dv = \sqrt{C} dz$, yields \begin{equation} I = \int \frac{\sqrt{C} dz}{ z^2 + 1} = \sqrt{C} \tan^{-1} z + C' \end{equation}
Now comes some algebra work. We need to do back-substitution to get the expression in terms of the original variable $\theta$.
Here is the back substitution:
\begin{eqnarray} I &=& \int \frac{\sqrt{C} dz}{ z^2 + 1} \\ &=& \sqrt{C} \tan^{-1} z + C' \\ &=& \sqrt{C} \tan^{-1} (v/\sqrt{C}) + C' \\ &=& \sqrt{C} \tan^{-1} (\sqrt{(k u^2 -1)/C}) + C' \\ &=& \sqrt{C} \tan^{-1} (\sqrt{k/t^2 -1/C}) + C' \\ &=& \sqrt{C} \tan^{-1} (\sqrt{[(1-C)/\cos^2 \theta -1]/C}) + C' \\ &=& \sqrt{C} \tan^{-1} \left ( \sqrt{ (\sin^2 \theta - C)/\cos^2 \theta /C} \right ) + C' \\ &=& \sqrt{C} \tan^{-1} \left ( \frac{\sqrt{\sin^2 \theta - C}}{\sqrt{C} \cos \theta} \right ) + C' \end{eqnarray}