Trig Integral Inequality Show that if $f$ is Riemann integrable on $[a,b]$, then 
$$\left(\int_{a}^{b}f(x)\sin x\ dx\right)^2+\left(\int_{a}^{b}f(x)\cos x\ dx\right)^2\le(b-a)\int_{a}^{b}f^2(x)\ dx.$$ 
I know I need to use the Hölder Inequality in some form, but I am not sure how to rewrite the left side of the equation. I know I can rewrite the right side as $\left(\int_{a}^{b}\sin^2x+\cos^2x\ dx\right)\left(\int_{a}^{b}f^2(x)\ dx\right)$, but I'm struggling to invoke any established inequalities because of the addition. Thanks for any shared insights!   
 A: By C-S $$(b-a)\int\limits_a^bf(x)^2dx=\int\limits_a^bf(x)^2\sin^2xdx\int_a^b1dx+\int\limits_a^bf(x)^2\cos^2xdx\int_a^b1dx\geq$$
$$\geq\left(\int\limits_a^bf(x)\sin{x}dx\right)^2+\left(\int\limits_a^bf(x)\cos{x}dx\right)^2.$$
Actually, $f^2(x)=f(f(x))$ and $f(x)^2=\left(f(x)\right)^2.$
A: 1.Prove that $<f,g>:=\int_a^bf(x)g(x)dx$ is an inner product. 
2.Observe that the norm it induces is $\|f\|=\big{(}\int_a^bf^2(x)dx\big{)}^{1/2}$.
3.Rewrite your inequality as $<f,\sin>^2+<f,\cos>^2\leq(b-a)\cdot\|f\|^2$
4.Using Cauchy-Schwarz, show that $<f,\sin>^2+<f,\cos>^2 \leq \|f\|^2(\|\sin\|^2+\|\cos\|^2)$
5.Observe that $\|\sin\|^2+\|\cos\|^2=\int_a^b\sin(x)^2dx+\int_a^b\cos(x)^2dx=\int_a^b1dx=b-a$.
A: $$\left(\int_{a}^{b}f(x)\sin x\ dx\right)^2+\left(\int_{a}^{b}f(x)\cos x\ dx\right)^2=\Big|\int^b_a e^{ix} f(x)\,dx\Big|^2$$
An application of Cauchy-Bunyakovsky's inequality yields
$$\Big|\int^b_a e^{ix} f(x)\,dx\Big|\leq\Big(\int^b_a|e^{2ix}|^2\,dx\Big)^{1/2}\Big(\int^b_a|f(x)|^2\,dx\Big)^{1/2}=(b-a)^{1/2}\int^b_a|f(x)|^2\,dx$$
