Solving the integral $\int_0^{\pi/2} \frac{\sin((2n+1)t)}{\sin t} \mathrm{d}t$ I really don't know how to solve this integral
$$\int_0^{\pi/2} \frac{\sin((2n+1)t)}{\sin t} \mathrm{d}t$$
Should I use firstly a formula of $\sin(a+b)$?
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\begin{align}
&\color{#00f}{\large\int_{0}^{\pi/2}{\sin\pars{\bracks{2n + 1}t} \over \sin\pars{t}}\,\dd t}
=\int_{0}^{\pi/2}{\expo{\ic\pars{2n + 1}t} - \expo{-\ic\pars{2n + 1}t}
\over \expo{\ic t} - \expo{-\ic t}}\,\dd t
\\[3mm] = & \ \int_{0}^{\pi/2}\expo{-2nt\,\ic}{\expo{\ic\pars{2n + 1}2t} - 1
\over \expo{2\ic t} - 1}\,\dd t
=\int_{0}^{\pi/2}\expo{-2nt\,\ic}\sum_{k = 0}^{2n}\expo{2\ic k t}\,\dd t
\\[3mm] = & \ \sum_{k = -n}^{n}\int_{0}^{\pi/2}\expo{2kt\,\ic}\,\dd t
= \sum_{k = -n}^{-1}\int_{0}^{\pi/2}\expo{2kt\,\ic}\,\dd t + {\pi \over 2}
+ \sum_{k = 1}^{n}\int_{0}^{\pi/2}\expo{2kt\,\ic}\,\dd t
\\[3mm] = &\ {\pi \over 2}
+ 2\sum_{k = 1}^{n}\
\overbrace{\int_{0}^{\pi/2}\cos\pars{2kt}\,\dd t}^{\ds{=\ 0}}
= \color{#00f}{\Large{\pi \over 2}}
\end{align}
A: This is known as the Dirichlet integral and the integrand sin(2n+1)t/sint = 1 + 2cos2t + 2cos4t + ...+ 2cos(2nt). You can use this to integrate term by term and the answer is: pi/2
A: It is a general very useful fact that $$D_n(t)=\sum_{k=-n}^ne^{ikt}=1+2\sum_{k=1}^n\cos kt=\frac{\sin\left(n+\frac 1 2\right)t}{\sin \frac t 2}$$
You can go head and try to prove it. $D_n(t)$ is known as the Dirichlet kernel, and is widely used in Fourier Theory. In particular, if $S_n(t)$ is the $n$-th partial sum of $f$'s  Fourier series, $$S_n(t)=\frac{1}{2\pi}\int_0^{2\pi}f(u)D_n(u-t)dt=(f\ast D_n)(t)$$
A: If $\displaystyle I_n=\int_0^{\dfrac\pi2} \frac{\sin((2n+1)t)}{\sin t} \mathrm{d}t$
$\displaystyle I_n-I_{n-1}=\int_0^{\dfrac\pi2} \frac{\sin((2n+1)t)-\sin((2n-1)t)}{\sin t} \mathrm{d}t$
Using Prosthaphaeresis Formulas, $\displaystyle \sin((2n+1)t)-\sin((2n-1)t)=2\sin t\cos2nt $
$\displaystyle\implies I_n-I_{n-1}=2\int_0^{\dfrac\pi2}\cos2nt\mathrm{d}t=\cdots =0$
Now, $\displaystyle I_0=\int_0^{\dfrac\pi2} \frac{\sin((2\cdot+1)t)}{\sin t} \mathrm{d}t=\int_0^{\dfrac\pi2}\mathrm{d}t=\cdots$
