Evaluation of $$\displaystyle \int_{-k}^{k}\frac{1}{\sqrt{\cos x-\cos k}}dx\;,$$

$\bf{My\; Try::}$ Let $$\displaystyle I = \int_{-k}^{k}\frac{1}{\sqrt{\cos x-\cos k}}dx = 2\int_{0}^{k}\frac{1}{\sqrt{\cos x-\cos k}}dx$$

Now Substiute $$\displaystyle \cos x = \frac{1-\tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}\;,$$ So we get $$\displaystyle I = 2\int_{0}^{k}\frac{\sec \frac{x}{2}}{\sqrt{1-\tan^2 \frac{x}{2}-\cos k(1+\tan^2 \frac{x}{2})}}dx$$

So we get $$\displaystyle I = 2\int_{0}^{k}\frac{\sec \frac{x}{2}}{\sqrt{(1-\cos k)-(1+\cos k)\tan^2\frac{x}{2}}}dx = 2\int_{0}^{k}\frac{\sec \frac{x}{2}}{\sqrt{2\sin^2\frac{k}{2}-2\cos^2 \frac{k}{2}\tan^2 \frac{x}{2}}}dx$$

Now How can I solve after that, Help me


  • $\begingroup$ u will need elliptic functions for this one $\endgroup$ – tired Nov 24 '15 at 15:46
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    $\begingroup$ $$\int_{-k}^{k}\frac{1}{\sqrt{\cos(x)-\cos(k)}}\space\text{d}x=$$ $$\frac{4\text{F}\left(\frac{k}{2}|\csc^2\left(\frac{k}{2}\right)\right)}{\sqrt{1-\cos(k)}}\space,\Re(k)>0\space\space\text{&&}\space\space\Im(k)=0$$ So: $$k\in\mathbb{R^+}$$ $\endgroup$ – Jan Nov 24 '15 at 18:46

I am trying to continue from your method but before that let me post my method:

use substitution $$\sqrt{cosx-cosk}=t$$ so $$\frac{dx}{\sqrt{cosx-cosk}}=\frac{-2dt}{sinx}=\frac{-2dt}{\sqrt{1-(t^2+cosk)^2}}=\frac{-2dt}{\sqrt{1-(t^2+cosk)}\sqrt{1+(t^2+cosk)}}=\frac{-2dt}{\sqrt{2sin^2\frac{k}{2}-t^2}\sqrt{2cos^2\frac{k}{2}+t^2}}$$

if we let $a=\sqrt{2}sin\frac{k}{2}$ and $b=\sqrt{2}cos\frac{k}{2}$ we have

$$I=\int\frac{-2dt}{\sqrt{a^2-t^2}\sqrt{b^2+t^2}}$$ which wolfram gives as

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    $\begingroup$ as long as tge square roots stay real u could simplify ur last equality nicely $\endgroup$ – tired Nov 24 '15 at 17:36

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