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?

## 1 Answer

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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 $p+q$.