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

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

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

$$\int\sqrt{1 + \cos^2(x)}\ \text{d}x = \sqrt{2}\ \text{E}\left(x, \frac{1}{2}\right)$$
Where $E$ stands for the elliptic integral of Second Kind.