nice exercise asked by Kunihiko Chikaya‎ in Enjoy Solving mathematics group

Exercise:

Compute, $\int_{-\pi }^{\pi }{\frac{\sin \left( n+\frac{1}{2} \right)\theta }{2\sin \frac{\theta }{2}}d\theta }$

Solution: Let ${{I}_{n}}=\int_{-\pi }^{\pi }{\frac{\sin \left( n+\frac{1}{2} \right)\theta }{2\sin \frac{\theta }{2}}d\theta \Leftrightarrow {{I}_{n+1}}=\int_{-\pi }^{\pi }{\frac{\sin \left( n+\frac{3}{2} \right)\theta }{2\sin \frac{\theta }{2}}d\theta }}$

But $\sin \left( a \right)-\sin \left( b \right)=2\cos \frac{a+b}{2}\sin \frac{a-b}{2}$

So $\sin \left( n+\frac{3}{2} \right)\theta -\sin \left( n+\frac{1}{2} \right)\theta =2\cos \frac{\left( n+n \right)\theta +\left( \frac{3}{2}+\frac{1}{2} \right)\theta }{2}\sin \frac{\left( n-n \right)\theta +\left( \frac{3}{2}-\frac{1}{2} \right)\theta }{2}$

                                        \[=2\cos \frac{\left( 2n+2 \right)\theta }{2}\sin \frac{\theta }{2}=2\cos \left( \left( n+1 \right)\theta  \right)\sin \frac{\theta }{2}\]

Hence ${{I}_{n+1}}-{{I}_{n}}=\int_{-\pi }^{\pi }{\frac{\sin \left( n+\frac{3}{2} \right)\theta }{2\sin \frac{\theta }{2}}d\theta }-\int_{-\pi }^{\pi }{\frac{\sin \left( n+\frac{1}{2} \right)\theta }{2\sin \frac{\theta }{2}}d\theta =}\int_{-\pi }^{\pi }{\frac{2\cos \left( \left( n+1 \right)\theta  \right)\sin \frac{\theta }{2}}{2\sin \frac{\theta }{2}}d\theta }$

Thus ${{I}_{n+1}}-{{I}_{n}}=\int_{-\pi }^{\pi }{\cos \left( \left( n+1 \right)\theta  \right)d\theta }$

Let $w=\left( n+1 \right)\theta \Leftrightarrow dw=\left( n+1 \right)d\theta \Leftrightarrow \frac{dw}{n+1}=d\theta $

So $\int_{-\pi }^{\pi }{\cos \left( \left( n+1 \right)\theta  \right)d\theta =\frac{1}{n+1}\int_{-\left( n+1 \right)\pi }^{\left( n+1 \right)\pi }{\cos wdw}=\frac{1}{n+1}\left( \sin w \right)_{-\left( n+1 \right)\pi }^{\left( n+1 \right)\pi }}$

                                $=\frac{1}{n+1}\left[ \sin \left( n+1 \right)\pi -\sin \left( -n-1 \right)\pi  \right]=0$


Therefore, ${{I}_{n+1}}={{I}_{n}}=....={{I}_{1}}={{I}_{0}}=\int_{-\pi }^{\pi }{\frac{\sin \frac{\theta }{2}}{2\sin \frac{\theta }{2}}d\theta =\int_{-\pi }^{\pi }{\frac{1}{2}d\theta }=\frac{\pi }{2}+\frac{\pi }{2}=\pi }$




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The idea of solution credit to Sofiane Ben Fredj