Given $\theta\in [0,2\pi)$ and let $a,b$ be two nonnegative integers. Consider the following integral:
$$I=\int\_0^{\pi}\sin^{2a+1}x\sin^{2b+1}(x+\theta)dx.$$
Since 
$$\int\_0^{\pi}\sin^mx\cos^{2n+1}xdx=0$$
for all nonnegative integers $m,n$, we have
\begin{eqnarray\*}
&&\int\_0^{\pi}\sin^{2a+1}x\sin^{2b+1}(x+\theta)dx\\
&=&\int\_0^{\pi}\sin^{2a+1}x(\sin x\cos\theta+\cos x\sin\theta)^{2b+1}dx\\
&=&\sum\_{k=0}^b\left(\binom{2b+1}{2k+1}\int\_0^{\pi}\sin^{2b-2k}\theta\sin^{2a+2k+2}x\cos^{2b-2k}xdx\right)(\cos\theta)^{2k+1}. 
\end{eqnarray\*}
Then the sign of $I$ is equal to the sign of $\cos\theta$. Is there any more direct or more simple way to see this?



The sign of the following integral
Replace $x$ to $\pi-x$, write $2I=2\int\_0^\pi f(x)dx=\int\_0^\pi f(x)+f(\pi-x)dx$ and note that


1) the sign of $\sin(x+\theta)+\sin(\pi-x+\theta)=2\sin x\cos \theta$ is the same as the sign of $\cos \theta$;


2) the sign of $p^{2b+1}+q^{2b+1}$ is the same as the sign of $p+q$. 


