Evaluation of $\displaystyle \int \sqrt{1+\cos^2 x}\,dx$

$\bf{My\; Try::}$ Let $$I = \int \sqrt{1+\cos^2 x},dx$$

Put $\cos x= t\;,$ Then $\displaystyle \,dx = -\frac{1}{\sin x}dx = -\frac{1}{\sqrt{1-t^2}}\,dt$

So $$I = -\int\sqrt{\frac{1+t^2}{1-t^2}}\,dt$$

Now How can i solve it after that, Help required, Thanks

  • 1
    $\begingroup$ en.wikipedia.org/wiki/… Doesn't look promising $\endgroup$
    – b00n heT
    Oct 29, 2016 at 17:10
  • $\begingroup$ This is an elliptic integral, you will not find an elementary antiderivative. $\endgroup$
    – user275377
    Oct 29, 2016 at 17:12
  • $\begingroup$ integral-calculator.com $\endgroup$
    – miracle173
    Jan 9, 2022 at 9:45

1 Answer 1


That is not an evaluable integral. In terms of elementary functions I mean, unless you consider Elliptic integrals as elementary.

First of all, we cannot even try any numerical or Series evaluation since the integral is unbounded. Your way of substitution is good, because as other users have already pointed out, the result is in terms of Special functions called the Elliptic Functions. That is:

$$\int\sqrt{1 + \cos^2(x)}\ \text{d}x = \sqrt{2}\ \text{E}\left(x, \frac{1}{2}\right)$$

Where $E$ stands for the elliptic integral of Second Kind.

More here: https://en.wikipedia.org/wiki/Elliptic_integral


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .