Elliptic integral

Exercise: (Elliptic Integrals )

Integrate , $\int_{1}^{\infty }{\frac{dx}{\sqrt{\left( {{x}^{2}}-2x+10 \right)\left( {{x}^{2}}-2x+7 \right)}}}$

Solution: we have ${{x}^{2}}-2x+10={{x}^{2}}-2x+1-1+10={{\left( x-1 \right)}^{2}}+9$

Also we have ${{x}^{2}}-2x+7={{x}^{2}}-2x+1-1+7={{\left( x-1 \right)}^{2}}+6$

So $\int_{1}^{\infty }{\frac{dx}{\sqrt{\left( {{x}^{2}}-2x+10 \right)\left( {{x}^{2}}-2x+7 \right)}}}=\int_{1}^{\infty }{\frac{dx}{\sqrt{{{\left( x-1 \right)}^{2}}+9}\times \sqrt{{{\left( x-1 \right)}^{2}}+6}}}$

Take $w=x-1\Leftrightarrow dw=dx$ & $w\left( 1 \right)=0\,\,,\,w\left( \infty  \right)=\infty $

So $\int_{1}^{\infty }{\frac{dx}{\sqrt{{{\left( x-1 \right)}^{2}}+9}\times \sqrt{{{\left( x-1 \right)}^{2}}+6}}=\int_{0}^{\infty }{\frac{dw}{\sqrt{{{w}^{2}}+9}\times \sqrt{{{w}^{2}}+6}}}}$

Let $w=\sqrt{6}\tan \theta \Rightarrow \theta =\arctan w\,\Leftrightarrow \theta =0\,\,\And \,\,\theta =\frac{\pi }{2}$ whenever $w=1\,\And \,w=\infty $

So $\int_{0}^{\infty }{\frac{dw}{\sqrt{{{w}^{2}}+9}\times \sqrt{{{w}^{2}}+6}}=\int_{0}^{\frac{\pi }{2}}{\frac{\sqrt{6}\,\,d\left( \tan \theta  \right)}{\sqrt{6{{\tan }^{2}}\theta +9}\times \sqrt{6{{\tan }^{2}}\theta +6}}}}$

but $d\left( \tan \theta  \right)={{\sec }^{2}}\theta \,d\theta =\left( 1+{{\tan }^{2}}\theta  \right)d\theta $

So $\int_{0}^{\frac{\pi }{2}}{\frac{\sqrt{6}d\left( \tan \theta  \right)}{\sqrt{6{{\tan }^{2}}\theta +9}\times \sqrt{6{{\tan }^{2}}\theta +6}}}$ $=\int_{0}^{\frac{\pi }{2}}{\frac{\sqrt{6}{{\sec }^{2}}\theta d\theta }{\sqrt{\left( 6{{\tan }^{2}}\theta +9 \right)6{{\sec }^{2}}\theta }}}$

$=\int_{0}^{\frac{\pi }{2}}{\frac{\sqrt{6}{{\sec }^{2}}\theta d\theta }{\sec \theta \sqrt{6}\sqrt{\left( 6{{\tan }^{2}}\theta +9 \right)}}}=\int_{0}^{\frac{\pi }{2}}{\frac{\sec \theta d\theta }{\sqrt{6{{\tan }^{2}}\theta +9}}}=\int_{0}^{\frac{\pi }{2}}{\frac{\frac{1}{\cos \theta }d\theta }{\sqrt{\frac{6{{\sin }^{2}}\theta +9{{\cos }^{2}}\theta }{{{\cos }^{2}}\theta }}}}$

$=\int_{0}^{\frac{\pi }{2}}{\frac{d\theta }{\sqrt{6{{\sin }^{2}}\theta +9{{\cos }^{2}}\theta }}}$ but ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \Leftrightarrow 9{{\cos }^{2}}\theta =9-9{{\sin }^{2}}\theta $

$=\int_{0}^{\frac{\pi }{2}}{\frac{d\theta }{\sqrt{6{{\sin }^{2}}\theta +9-9{{\sin }^{2}}\theta }}=\int_{0}^{\frac{\pi }{2}}{\frac{d\theta }{\sqrt{9-3{{\sin }^{2}}\theta }}}=\int_{0}^{\frac{\pi }{2}}{\frac{d\theta }{\sqrt{9\left( 1-\frac{1}{3}{{\sin }^{2}}\theta  \right)}}}}$

$=\int_{0}^{\frac{\pi }{2}}{\frac{d\theta }{3\sqrt{1-\frac{1}{3}{{\sin }^{2}}\theta }}}=\frac{1}{3}\int_{0}^{\frac{\pi }{2}}{\frac{d\theta }{\sqrt{1-\frac{1}{3}{{\sin }^{2}}\theta }}}=\frac{1}{3}EllipticF\left( \theta=\frac{\pi}{2} ,\frac{1}{\sqrt{3}} \right)$