Nice integral exercise asked by Dan Sitaru in the calculus4u group

Exercise:

Compute, $\int_{n}^{n+1}{\frac{\sqrt{n+1-x}}{\sqrt{n+1-x}+\sqrt{x-n}{{e}^{2x-2n-1}}}dx}$

Solution: Let $x-n={{\sin }^{2}}\theta \Leftrightarrow dx=2\sin \theta \cos \theta d\theta $

If $x=n\Leftrightarrow {{\sin }^{2}}\theta =0\Leftrightarrow \theta =0$ & if $x=n+1\Leftrightarrow {{\sin }^{2}}\theta =1\Leftrightarrow \theta =\frac{\pi }{2}$

So $\int_{n}^{n+1}{\frac{\sqrt{n+1-x}}{\sqrt{n+1-x}+\sqrt{x-n}{{e}^{2x-2n-1}}}dx}=\int_{0}^{\frac{\pi }{2}}{\frac{2\sin \theta \cos \theta \sqrt{1-{{\sin }^{2}}\theta }}{\sqrt{1-{{\sin }^{2}}\theta }+\sqrt{{{\sin }^{2}}\theta }{{e}^{2\left( {{\sin }^{2}}\theta  \right)-1}}}d\theta }$

$=\int_{0}^{\frac{\pi }{2}}{\frac{2\sin \theta {{\cos }^{2}}\theta }{\cos \theta +\sin \theta {{e}^{2{{\sin }^{2}}\theta -1}}}d\theta }=\int_{0}^{\frac{\pi }{2}}{\frac{\sin 2\theta \cos \theta }{\cos \theta +\sin \theta {{e}^{-\cos 2\theta }}}d\theta }$

$=\int_{0}^{\frac{\pi }{2}}{\frac{2\sin \left( \frac{\pi }{2}-\theta  \right)\cos \left( \frac{\pi }{2}-\theta  \right)\cos \left( \frac{\pi }{2}-\theta  \right)}{\cos \left( \frac{\pi }{2}-\theta  \right)+\sin \left( \frac{\pi }{2}-\theta  \right){{e}^{2{{\sin }^{2}}\left( \frac{\pi }{2}-\theta  \right)-1}}}d\theta }$

$=\int_{0}^{\frac{\pi }{2}}{\frac{\sin 2\theta \sin \theta }{\sin \theta +\cos \theta {{e}^{\cos 2\theta }}}\times \frac{{{e}^{-\cos 2\theta }}}{{{e}^{-\cos 2\theta }}}d\theta }=\int_{0}^{\frac{\pi }{2}}{\frac{{{e}^{-\cos 2\theta }}\sin 2\theta \sin \theta }{{{e}^{-\cos 2\theta }}\sin \theta +\cos \theta }d\theta }$

So $2\int_{0}^{\frac{\pi }{2}}{\frac{\sin 2\theta \cos \theta }{\cos \theta +\sin \theta {{e}^{-\cos 2\theta }}}d\theta }=\int_{0}^{\frac{\pi }{2}}{\frac{\sin 2\theta \left( \cos \theta +\sin \theta {{e}^{-\cos 2\theta }} \right)}{\cos \theta +\sin \theta {{e}^{-\cos 2\theta }}}d\theta }=\int_{0}^{\frac{\pi }{2}}{\sin 2\theta }\,d\theta $

Thus $\int_{0}^{\frac{\pi }{2}}{\frac{\sin 2\theta \cos \theta }{\cos \theta +\sin \theta {{e}^{-\cos 2\theta }}}d\theta =\frac{1}{2}\int_{0}^{\frac{\pi }{2}}{\sin 2\theta }\,d\theta }=\frac{1}{2}\left( \frac{-1}{2}\cos 2\theta  \right)_{0}^{\frac{\pi }{2}}=\frac{1}{2}$

Therefore, $\int_{n}^{n+1}{\frac{\sqrt{n+1-x}}{\sqrt{n+1-x}+\sqrt{x-n}{{e}^{2x-2n-1}}}dx}=\frac{1}{2}$

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the idea of solution credit to Diego Alvariz