2
$\begingroup$

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

Thanks

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

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

enter image description here

$\endgroup$
  • 1
    $\begingroup$ as long as tge square roots stay real u could simplify ur last equality nicely $\endgroup$ – tired Nov 24 '15 at 17:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.