I'm trying to show that \begin{align} \left|\int_{-\pi}^\pi [f(t)\cos t - f'(t) \sin t]\, dt\right| \le \sqrt{2\pi}\left(\int_{-\pi}^\pi [|f(t)|^2 + |f'(t)|^2]\, dt\right)^{1/2}. \end{align}
However I do not see how to do this. If I use the triangle inequality I get \begin{align} \left|\int_{-\pi}^\pi [f(t)\cos t - f'(t) \sin t]\, dt\right| = \left| \int_{-\pi}^\pi f(t)\cos t\, dt - \int_{-\pi}^\pi f'(t)\sin t\, dt\right| \le \left| \int_{-\pi}^\pi f(t)\cos t\, dt\right| + \left|\int_{-\pi}^\pi f'(t)\sin t\, dt\right|. \end{align}
If I then use Schwarz inequality I get that $$ \left| \int_{-\pi}^\pi f(t)\cos t\, dt\right| \le \sqrt{\pi} \left(\int_{-\pi}^\pi |f(t)|^2dt\right)^{1/2}, $$ so that \begin{align} \left|\int_{-\pi}^\pi [f(t)\cos t - f'(t) \sin t]\, dt\right| \le \sqrt{\pi} \left(\int_{-\pi}^\pi |f(t)|^2dt\right)^{1/2} + \sqrt{\pi} \left(\int_{-\pi}^\pi |f'(t)|^2dt\right)^{1/2}. \end{align}
I feel close to the expected result, but don't know how to improve. Any help will be appreciated. Cheers
By Cauchy's inequality $$\vert f(t)\cos t-f'(t)\sin t \vert =\vert (f(t),f'(t))\cdot (\cos t,-\sin t) \vert \le \sqrt{(f(t))^2+(f'(t))^2} \sqrt{(\cos t)^2+(\sin t)^2} =\sqrt{(f(t))^2+(f'(t))^2}.$$ Then apply Holder's