Integral exercise asked by Dan Stirua in many math groups .

Exercise:

Compute, $\int_{2}^{4}{{{e}^{x}}\left( \arcsin \left( x-3 \right)+2\arctan \left( \sqrt{\frac{4-x}{x-2}} \right) \right)dx}$

Solution: Let $u=x-3\Leftrightarrow du=dx\,\,\And \,\,u\left( 2 \right)=-1\,\,,\,\,u\left( 4 \right)=1$

So $\int_{2}^{4}{{{e}^{x}}\left( \arcsin \left( x-3 \right)+2\arctan \left( \sqrt{\frac{4-x}{x-2}} \right) \right)dx}$

$=\int_{-1}^{1}{{{e}^{u+3}}\left( \arcsin u+2\arctan \left( \sqrt{\frac{4-u-3}{u+3-2}} \right) \right)du}={{e}^{3}}\int_{-1}^{1}{{{e}^{u}}\left( \arcsin u+2\arctan \left( \sqrt{\frac{1-u}{u+1}} \right) \right)du}$

Let $\arctan x=\theta \Leftrightarrow x=\tan \theta $ and we know that $\tan 2\theta =\frac{2\tan \theta }{1-{{\tan }^{2}}\theta }$

So $2\theta =\arctan \left( \frac{2x}{1-{{x}^{2}}} \right)$ i.e $2\arctan x=\arctan \left( \frac{2x}{1-{{x}^{2}}} \right)$ ….(*)

So $2\arctan \left( \sqrt{\frac{1-u}{u+1}} \right)=\arctan \left( \frac{2\sqrt{\frac{1-u}{u+1}}}{1-\frac{1-u}{u+1}} \right)=\arctan \left( \frac{\frac{2\sqrt{1-u}}{\sqrt{u+1}}}{\frac{u+1-1+u}{u+1}} \right)$
                          $=\arctan \left( \frac{\left( u+1 \right)\sqrt{1-u}}{u\sqrt{1+u}} \right)=\arctan \left( \frac{\sqrt{1-{{u}^{2}}}}{u} \right)$

Let $w\left( u \right)=\frac{u}{\sqrt{1-{{u}^{2}}}}$ and we know that$\arcsin u=\arctan \frac{u}{\sqrt{1-{{u}^{2}}}}\,\,\,,\,\,{{u}^{2}}\le 1$ , then

$f\left( u \right)=\arcsin u+2\arctan \left( \sqrt{\frac{1-u}{u+1}} \right)=\arctan \left( \frac{u}{\sqrt{1-{{u}^{2}}}} \right)+\arctan \left( \frac{\sqrt{1-{{u}^{2}}}}{u} \right)$

        $=\arctan \left( w\left( u \right) \right)+\arctan \left( \frac{1}{w\left( u \right)} \right)$ Notice that $\frac{d}{dx}\left( \arctan x \right)=\frac{1}{{{x}^{2}}+1}$

Then $f'\left( u \right)=\frac{1}{{{w}^{2}}+1}-\frac{1/{{w}^{2}}}{\frac{1}{{{w}^{2}}}+1}=\frac{1}{{{w}^{2}}+1}-\frac{1/{{w}^{2}}}{\frac{1+{{w}^{2}}}{{{w}^{2}}}}=\frac{1}{{{w}^{2}}+1}-\frac{1}{{{w}^{2}}+1}=0$

So by first derivative test we get $f=cst$ so $f\left( 1 \right)=f\left( 0 \right)=\frac{\pi }{2}$

Thus $\arcsin u+2\arctan \left( \sqrt{\frac{1-u}{u+1}} \right)=\frac{\pi }{2}\,\,\,\,\,,\,\,\,\,\,\forall \,\,\,\left| u \right|\le 1$

So $\int_{-1}^{1}{{{e}^{u+3}}\left[ \arcsin u+2\arctan \left( \sqrt{\frac{1-u}{u+1}} \right) \right]du}={{e}^{3}}\int_{-1}^{1}{{{e}^{u}}\left( \frac{\pi }{2} \right)du}=\frac{\pi {{e}^{3}}}{2}\left( {{e}^{u}} \right)_{-1}^{1}=\frac{\pi }{2}\left( {{e}^{4}}-{{e}^{2}} \right)$

Remark:

We can prove that , $\arctan u+\arctan \frac{1}{u}=\frac{\pi }{2}$ by taking $\theta =\arctan u\,\,\And \,\,\alpha =\arctan \frac{1}{u}$ , then

$\tan \theta =u\,\,\And \,\,\tan \alpha =\frac{1}{u}\Leftrightarrow \tan \theta \tan \alpha =1$ and

we know that , $\tan \left( \alpha +\theta  \right)=\frac{\tan \alpha +\tan \theta }{1-\tan \theta \tan \alpha }=\frac{\tan \alpha +\tan \theta }{0}=\infty =\tan \frac{\pi }{2}$

thus $\alpha +\theta =\frac{\pi }{2}\,\,\Leftrightarrow \arctan u+\arctan \frac{1}{u}=\frac{\pi }{2}$