Integral Exercise solved by using Elliptic integral Method asked by Katrina Linda in the Pretty Math group

Exercise:

Compute, $\int_{0}^{1}{\frac{dx}{\sqrt{1+{{x}^{2}}}\sqrt{1+2{{x}^{2}}}}}$

Let $x=\tan \theta \Leftrightarrow dx={{\sec }^{2}}\theta \,d\theta $

So $\int_{0}^{1}{\frac{dx}{\sqrt{1+{{x}^{2}}}\sqrt{1+2{{x}^{2}}}}=\int_{0}^{\frac{\pi }{4}}{\frac{{{\sec }^{2}}\theta }{\sqrt{1+{{\tan }^{2}}\theta }\sqrt{1+2{{\tan }^{2}}\theta }}d\theta }}$

$=\int_{0}^{\frac{\pi }{4}}{\frac{{{\sec }^{2}}\theta }{\sqrt{{{\sec }^{2}}\theta }\sqrt{1+2{{\tan }^{2}}\theta }}d\theta }=\int_{0}^{\frac{\pi }{4}}{\frac{\sec \theta }{\sqrt{1+2{{\tan }^{2}}\theta }}d\theta }$

But $\frac{\sec \theta }{\sqrt{1+2{{\tan }^{2}}\theta }}=\frac{\frac{1}{\cos \theta }}{\sqrt{1+2\frac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }}}=\frac{\frac{1}{\cos \theta }}{\sqrt{\frac{{{\cos }^{2}}\theta +2{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }}}=\frac{1}{\sqrt{{{\cos }^{2}}\theta +2{{\sin }^{2}}\theta }}$

So $\int_{0}^{\frac{\pi }{4}}{\frac{\sec \theta }{\sqrt{1+2{{\tan }^{2}}\theta }}d\theta }=\int_{0}^{\frac{\pi }{4}}{\frac{d\theta }{\sqrt{{{\cos }^{2}}\theta +2{{\sin }^{2}}\theta }}=\int_{0}^{\frac{\pi }{4}}{\frac{d\theta }{\sqrt{1+{{\sin }^{2}}\theta }}}}=F\left[ \frac{\pi }{4},-1 \right]$

Since $\int_{0}^{\frac{\pi }{4}}{\frac{d\theta }{\sqrt{1+{{\sin }^{2}}\theta }}}=\int_{0}^{\frac{\pi }{4}}{\frac{d\theta }{\sqrt{1-\left( -{{\sin }^{2}}\theta  \right)}}=\int_{0}^{\frac{\pi }{4}}{\frac{d\theta }{\sqrt{1-k{{\sin }^{2}}\theta }}=F\left( \frac{\pi }{4},k \right)=F\left( \frac{\pi }{4},-1 \right)}}$