Evaluating trigonometric integrals of the form $\int \frac{C \; d \theta}{ \sin \theta \sqrt{\sin^2 \theta - C }}$

Question

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.

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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}

Answer 1

Let $\displaystyle I = \int\frac{Cd\theta}{\sin \theta \sqrt{\sin^2 \theta - C}} = \int\frac{C \sin \theta}{\sin^2 \theta \sqrt{\sin^2 \theta -C}}d\theta$

So we get $$I = \int\frac{C\sin \theta}{(1-\cos^2 \theta)\sqrt{1-\cos^2 \theta -C}}d\theta\;,$$

Now Put $\cos \theta = t\;,$ Then $\sin \theta d\theta = -dt$

so we get $$I = -\int\frac{C}{(1-t^2)\sqrt{\underbrace{(1-C)}_{k}-t^2}}dt = -C\int\frac{1}{(1-t^2)\sqrt{k-t^2}}dt$$

Now Put $\displaystyle t = \frac{1}{u}\;,$ Then $\displaystyle dt = -\frac{1}{u^2}du$

So we get $$I = C\int\frac{u}{(u^2-1)\sqrt{ku^2-1}}du$$

Now at last Put $(ku^2-1)=v^2\;,$ Then $\displaystyle udu = \frac{1}{k}vdv$

So we get $$I = \frac{C}{k}\int\frac{k}{v^2+1}dv = C\int\frac{1}{v^2+1}dv= C\tan^{-1}(v)+\mathcal{C'}$$

Answer 2

$$\int \frac{C \; d \theta}{ \sin \theta \sqrt{\sin^2 \theta - C }}$$ $$=\int \frac{C \; \csc ^2 \theta d \theta}{ \csc ^2 \theta \cdot \sin \theta \sqrt{\sin^2 \theta - C }}$$ $$=\int \frac{C \; \csc ^2 \theta d \theta}{ \sqrt{1 - C \csc^2 \theta}}$$ $$=\int \frac{C \; \csc ^2 \theta d \theta}{ \sqrt{1 - C (1+\cot^2 \theta)}}$$ Use $\sqrt{C} \cot \theta = z$ Then $-\sqrt{C} \; \csc ^2 \theta d \theta = dz$ $$=\int \frac{-\sqrt{C} dz}{ \sqrt{1 - C - z^2}}$$ Say $1-C=a^2$ $$=\int \frac{-\sqrt{C} dz}{ \sqrt{a^2 - z^2}}$$ $$=-\sqrt{C}\sin^{-1}\left(\frac{z}{a}\right)+k$$ $$=-\sqrt{C}\sin^{-1}\left(\frac{\sqrt{C} \cot \theta}{a}\right)+k$$

Answer 3

Hint One can handle the second form of the integral by writing it as $$\int \frac{x \,dx}{x^2 \sqrt{a + b x^2 + c x^4}}$$ and substituting $$u = x^2, \quad du = 2 x \,dx ,$$ to write the integral as $$\int \frac{du}{u \sqrt{a + b u + c u^2}} .$$ This latter integral is a standard sort of exercise and can be handled with trigonometric (or hyperbolic trigonometric) substitution. The appropriate substitution, as well as the qualitative behavior of the antiderivative, depends on the sign of the discriminant $b^2 - 4 c a$ of the quadratic expression in the radical. (See this integrable table.)