Problem source

Let $\mathrm{f}(x)=\dfrac{\sin(n+\frac{1}{2})x}{\sin\frac{1}{2}x}$ for $0 < x\leqslant\pi.$ 
\begin{questionparts}
\item Using the formula 
\[
2\sin\tfrac{1}{2}x\cos kx=\sin(k+\tfrac{1}{2})x-\sin(k-\tfrac{1}{2})x
\]
(which you may assume), or otherwise, show that 
\[
\mathrm{f}(x)=1+2\sum_{k=1}^{n}\cos kx\,.
\]
\item Find ${\displaystyle \int_{0}^{\pi}\mathrm{f}(x)\,\mathrm{d}x}$ and ${\displaystyle \int_{0}^{\pi}\mathrm{f}(x)\cos x\,\mathrm{d}x}.$ 
\end{questionparts}

Solution source

\begin{questionparts}
\item $\,$
\begin{align*}
&& 2\sin \tfrac12 x \sum_{k=1}^n \cos kx &= \sum_{k=1}^n 2\sin \tfrac12 x \cos kx \\
&&&= \sum_{k=1}^n \left ( \sin(k+\tfrac12)x - \sin(k - \tfrac12)x \right) \\
&&&=  \left ( \sin\tfrac32x - \sin\tfrac12x \right) + \\
&&&\quad  \quad \left ( \sin\tfrac52x - \sin \tfrac32 x \right) + \\
&&&\quad \quad \quad +\cdots + \\
&&&\quad \quad \quad \quad +\left ( \sin(n+\tfrac12)x - \sin(n - \tfrac12)x \right) \\
&&&= \sin(n+\tfrac12)x - \sin \tfrac12 x \\
\Rightarrow && \sin(n+\tfrac12)x &= \sin \tfrac12 x + 2\sin \tfrac12 x \sum_{k=1}^n \cos kx \\
\Rightarrow && f(x) &= 1 + 2 \sum_{k=1}^n \cos kx
\end{align*}

\item $\,$
\begin{align*}
&& \int_0^{\pi} f(x) \d x &= \int_0^{\pi} \left (1 + 2 \sum_{k=1}^n \cos kx \right) \d x \\
&&&= \pi + 2 \left [ \sum_{k=1}^n \frac{1}{k} \sin k x\right]_0^\pi \\
&&&= \pi \\
\\
&& \int_0^{\pi} f(x) \cos x  \d x &= \int_0^{\pi} \left (\cos x + 2 \sum_{k=1}^n \cos kx \cos x \right) \d x \\
&&&= 0 + \sum_{k=1}^n \left ( \int_0^{\pi} 2 \cos k x \cos x  \d x  \right) \\
&&&= \sum_{k=1}^n \left ( \int_0^{\pi} (\cos (k+1)x - \cos (k-1) x)\d x\right) \\
&&&= -\pi
\end{align*}

\end{questionparts}