integral exercise asked by Dan Sitaru in many math groups

Exercise:

Integrate, $\int{\frac{dx}{\left( 1+2016{{x}^{2}} \right)\sqrt{1+2015{{x}^{2}}}}}$

Solution: Let $x=\frac{\tan \theta }{\sqrt{2015}}\Leftrightarrow dx=\frac{{{\sec }^{2}}\theta }{\sqrt{2015}}d\theta $

So $\sqrt{1+2015{{x}^{2}}}=\sqrt{1+2015\frac{{{\tan }^{2}}\theta }{2015}}=\sqrt{1+{{\tan }^{2}}\theta }=\sec \theta $ and

$1+2016{{x}^{2}}=1+\frac{2016{{\tan }^{2}}\theta }{2015}$

$\int{\frac{dx}{\left( 1+2016{{x}^{2}} \right)\sqrt{1+2015{{x}^{2}}}}=}\int{\frac{1}{\sec \theta \left( 1+\frac{2016{{\tan }^{2}}\theta }{2015} \right)}\times \frac{{{\sec }^{2}}\theta }{\sqrt{2015}}d\theta }$

                                      $=\frac{1}{\sqrt{2015}}\int{\frac{\sec \theta }{1+\frac{2016{{\tan }^{2}}\theta }{2015}}d\theta }$

$\frac{\sec \theta }{1+\frac{2016}{2015}{{\tan }^{2}}\theta }=\frac{\frac{1}{\cos \theta }}{1+\frac{2016}{2015}\times \frac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }}=\frac{\frac{1}{\cos \theta }}{\frac{2015{{\cos }^{2}}\theta +2016{{\sin }^{2}}\theta }{2015{{\cos }^{2}}\theta }}$

$=\frac{2015{{\cos }^{2}}\theta }{\cos \theta \left( 2015{{\cos }^{2}}\theta +2016{{\sin }^{2}}\theta  \right)}=\frac{2015\cos \theta }{2015+{{\sin }^{2}}\theta }$

So $\frac{1}{\sqrt{2015}}\int{\frac{\sec \theta }{1+\frac{2016{{\tan }^{2}}\theta }{2015}}d\theta =\frac{1}{\sqrt{2015}}\int{\frac{2015\cos \theta }{2015+{{\sin }^{2}}\theta }d\theta }=\frac{2015}{\sqrt{2015}}\int{\frac{\cos \theta }{2015+{{\sin }^{2}}\theta }d\theta }}$

Let $w=\sin \theta \Leftrightarrow dw=\cos \theta \,d\theta $ so $\frac{2015}{\sqrt{2015}}\int{\frac{\cos \theta d\theta }{2015+{{\sin }^{2}}\theta }}=\frac{2015}{\sqrt{2015}}\int{\frac{dw}{2015+{{w}^{2}}}}$

Let $w=\sqrt{2015}t\Leftrightarrow dw=\sqrt{2015}dt$

So $\frac{2015}{\sqrt{2015}}\int{\frac{dw}{2015+{{w}^{2}}}}=\frac{2015}{\sqrt{2015}}\int{\frac{\sqrt{2015}dt}{2015\left( 1+{{t}^{2}} \right)}=\int{\frac{dt}{1+{{t}^{2}}}=\arctan \left( t \right)+c}}$

$=\arctan \left( \frac{w}{\sqrt{2015}} \right)+c=\arctan \left( \frac{\sin \theta }{\sqrt{2015}} \right)+c$

We have $x=\frac{\tan \theta }{\sqrt{2015}}\Leftrightarrow {{x}^{2}}=\frac{{{\tan }^{2}}\theta }{2015}\Leftrightarrow 2015{{x}^{2}}={{\sec }^{2}}\theta -1\Leftrightarrow 1+2015{{x}^{2}}={{\sec }^{2}}\theta $

$\Leftrightarrow \sec \theta =\sqrt{1+2015{{x}^{2}}}\Leftrightarrow \frac{1}{\cos \theta }=\sqrt{1+2015{{x}^{2}}}\Leftrightarrow \cos \theta =\frac{1}{\sqrt{1+2015{{x}^{2}}}}\Leftrightarrow {{\cos }^{2}}\theta =\frac{1}{1+2015{{x}^{2}}}$

$\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }=\sqrt{1-\frac{1}{1+2015{{x}^{2}}}}=\sqrt{\frac{1+2015{{x}^{2}}-1}{1+2015{{x}^{2}}}}=\sqrt{\frac{2015{{x}^{2}}}{1+2015{{x}^{2}}}}=\frac{x\sqrt{2015}}{\sqrt{1+2015{{x}^{2}}}}$

Thus $\int{\frac{dx}{\left( 1+2016{{x}^{2}} \right)\sqrt{1+2015{{x}^{2}}}}=\arctan \left( \frac{x}{\sqrt{1+2015{{x}^{2}}}} \right)+c}$