How will you approach this?

Let us rewrite the sine functions using Euler's identity $\sin x = \cfrac{\mathrm{e}^{ix} - \mathrm{e}^{-ix}}{2i} = \cfrac{\mathrm{e}^{2ix} - 1}{2i\mathrm{e}^{ix}}$ (the latter received by multiplying the numerator and the denominator by $e^{ix}$ );

Step 1: $\displaystyle\int _{ 0 }^{ \pi }{ \cfrac { \sin\left( n+\cfrac { 1 }{ 2 } \right) x }{ \sin(\frac {x}{2} ) } } \mathrm {d} x = \int _{0}^{\pi}{\cfrac{\mathrm{e}^{(2n+1)ix} - 1}{2i\mathrm{e}^{(n+\frac{1}{2})ix}} \cfrac{2i\mathrm{e}^{\frac{ix}{2}}}{\mathrm{e}^{ix} - 1} \mathrm{d}x}$ ;

Step 2: The $2i$ 's cancel out and we use a property of exponents to receive $\displaystyle\int _{0}^{\pi}{\cfrac{(\mathrm{e}^{-ixn})(\mathrm{e}^{(2n+1)ix} - 1)}{\mathrm{e}^{ix} - 1} \mathrm{d}x}$

The integrand resembles a formula for the sum of geometric series up to the $n$ -th term: $S_n = \sum\limits_{i=1}^n b_i = \cfrac{ b_1 (1 - q^{n} ) }{ 1-q }$ , where $b_1$ is the first term of the series and $q$ is its common ratio. In this case we have the sum of geometric series up to the $2n+1$ -th term with the first term being $\mathrm{e}^{-ixn}$ and the common ratio being $e^{ix}$ ;

Step 3: We rewrite the geometric series as a sum to get $\displaystyle\int_{0}^{\pi}{ \sum\limits_{k=1}^{2n+1} \mathrm{e}^{-ix(n-k+1)}} \mathrm{d}x = \displaystyle\int _{0}^{\pi}{(\mathrm{e}^{-ixn} + \mathrm{e}^{-ix(n-1)} + \dots + \mathrm{e}^{ix(n-1)} + \mathrm{e}^{ixn})} \mathrm{d}x$

Step 4: To deal with this one, we need to use Euler's identity for the cosine function: $\cos x = \cfrac{\mathrm{e}^{ix} + \mathrm{e}^{-ix}}{2}$ ; we also have to keep in mind that one of the terms is $\mathrm{e}^{-ix(n-n)} = \mathrm{e}^0 = 1$ $\displaystyle\int _{0}^{\pi}{(2\cos{nx} + 2\cos{(n-1)}x + \dots + 2\cos{x} + 1 )} \mathrm{d}x$

Step 5: We use the sum rule for definite integrals to separate each term: $\displaystyle\int _{0}^{\pi}{2\cos{nx}} \mathrm{d}x + \displaystyle\int _{0}^{\pi}{2\cos{(n-1)x}} \mathrm{d}x + \dots + \displaystyle\int _{0}^{\pi}{2\cos{x}} \mathrm{d}x + \displaystyle\int_{0}^{\pi} 1 \mathrm{d}x$

Now instead of integrating every cosine, notice that we can rewrite any of them as $2\cos{kx}$ where $k$ is a natural number from 1 to $n$ since $n$ is natural.

$\displaystyle\int_0^{\pi}{2\cos{kx}} \mathrm{d}x = \frac{2}{k} \sin{kx}|_0^{\pi} = \frac {2}{k}\sin{\pi k} - \frac {2}{k} \sin{0} = 0$

This means that all of the integrals which include cosines equal zero. Now we are only left with $\displaystyle\int_{0}^{\pi} 1 \mathrm{d}x$ = $\pi$ which is the answer.