Let $x\in C^1[-\pi,\pi].$ Prove that $|\int_{-\pi}^{\pi}(x(t).cost-x'(t).sint)dt|^2\le 2\pi\int_{-\pi}^{\pi}(|x(t)|^2+|x'(t)|^2)dt$

Question

Let $x\in C^1[-\pi,\pi].$ Prove that $|\int_{-\pi}^{\pi}(x(t).cost-x'(t).sint)dt|^2\le 2\pi\int_{-\pi}^{\pi}(|x(t)|^2+|x'(t)|^2)dt$

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Answer 1

\begin{align}&\left|\int_{-\pi}^{\pi}(x(t).cost-x'(t).sint)dt\right|^2 \\&=\left|\int_{-\pi}^{\pi} \frac{(x(t).cost-x'(t).sint)}{\sqrt{|x(t)|^2+|x'(t)|^2}}.\sqrt{|x(t)|^2+|x'(t)|^2}dt\right|^2 \\&\le \int_{-\pi}^{\pi}\left|\frac{(x(t).cost-x'(t).sint)}{\sqrt{|x(t)|^2+|x'(t)|^2}}\right|^2 \,dt .\int_{-\pi}^{\pi}(|x(t)|^2+|x'(t)|^2)dt \\&\le 2\pi\int_{-\pi}^{\pi}(|x(t)|^2+|x'(t)|^2)dt\end{align}

The First Inequality is Cauchy Schwarz Inequality and the second is,

$$\left|\dfrac{(x(t).cost-x'(t).sint)}{\sqrt{|x(t)|^2+|x'(t)|^2}}\right| \le 1$$