This question is about exercise 19 in the third chapter of Stein and Shakarchi's Fourier analysis. We are asked to prove that $$ \label{eq:1} \sum_{n=1}^N \frac{\sin nx}{n}=\frac{1}{2}\int_0^x(D_N(t)-1)dt, \tag{*} $$ where $D_N(t) = \sum_{|n|<=N} \exp(int) = 1 + 2\sum_{n=1}^N \cos nt$ is the Dirichlet kernel. And then as a consequence of the identity in $\eqref{eq:1}$ to prove that $\sum_{n=1}^N \frac{\sin nx}{n}$ is uniformly bounded in $N$ and $x$ for $0 \leq x \leq \pi$ (and hence for all $x$ given that the expression in the LHS above is odd and $2\pi$ periodic).
The identity in $\eqref{eq:1}$ easily follows from the fact that $$ \frac{d}{dt} \sum_{n=1}^N \frac{\sin nt}{n} = \frac{1}{2}(D_N(t) - 1), $$ but I am unable to see how the integral representation implies uniform boundedness in $N$ and $x$.
Let $$F_N(x) = \int_0^x (D_N(t) - 1)dt = \int_0^x \left( \frac{ \sin (N+1/2)t}{\sin(t/2)} -1 \right) dt$$ and let $$G_N(t) = \int_0^x \left( \frac{\sin (N+1/2)t}{t/2} - 1 \right)dt.$$
We have to show $$ \sup_{ \substack{ 0 \leq x \leq \pi \\ N \geq 1} } |F_N(x)| < \infty.$$
It is sufficient to show $ \sup_{ \substack{ 0 \leq x \leq \pi \\ N \geq 1} } |F_N(x) - G_N(x)| < \infty$ and $ \sup_{ \substack{ 0 \leq x \leq \pi \\ N \geq 1} } |G_N(x)| < \infty$.
Note that $$g(x) = \frac{1}{\sin(x/2)} - \frac{1}{x/2}$$ is a continuously differentiable function for $x \in [0,\pi].$
Now $$F_N(x) - G_N(x) = \int_{0}^x \sin( (N+1/2) t) g(t) dt = \int_{0}^x \frac{d}{dt} \left(\frac{1 -\cos( (N+1/2) t)}{(N+1/2)}\right) g(t)dt . $$
Integrating by parts we get, $$ F_N(x) - G_N(x) = \frac{1-\cos( (N+\frac{1}{2})x)}{(N+\frac{1}{2})}g(x) - \frac{1}{(N+\frac{1}{2})}\int_{0}^x (1 - \cos( (N+\frac{1}{2})t )) g'(t) dt.$$
Let $M_1 = \sup_{0 \leq x \leq \pi} |g(x)|$ and let $M_2 = \sup_{0 \leq x \leq\pi} |g'(x)|,$ then $$ \label{e:1} |F_N(x) - G_N(x)| \leq \frac{2M_1}{N+\frac{1}{2}} + \frac{2M_2\pi}{N+\frac{1}{2}} \leq 4M_1 + 4M_2\pi \tag{*}$$ for all $N \geq 1$ and $0 \leq x \leq \pi.$
We have, $$ G_N(x) = 2\int_0^x \frac{\sin(N+\frac{1}{2})t}{t}dt - x = 2 \int_0^{ (N+1/2)x} \frac{\sin t}{t}dt - x. $$
From the convergence of the improper Riemann integral $ \int_0^\infty \frac{\sin t}{t}dt$ we have $ \sup_{0 \leq x < \infty} | \int_0^x \frac{\sin t}{t}dt |\, \colon= M_3 < \infty$.
From which it follows $\label{e:2}|G_N(x)| \leq 2 M_3 + \pi \tag{+}$ for all $N \geq 1$ and all $0 \leq x \leq \pi$.
From $\eqref{e:1}$ and $\eqref{e:2}$ the result follows.