# Compute $\int_{0}^{\frac \pi2} \sin((m-1)x)\sin(mx)\sin((m+1)x)dx$.

Let $m$ be a natural odd number. Compute: $$\int_{0}^{\frac \pi2} \sin((m-1)x)\sin(mx)\sin((m+1)x)dx.$$

I tried using the integral property:

$$\int_0^af(x)dx=\int_0^af(a-x)dx,$$

and then adding them up but it doesn't work... I'm thinking maybe I should group the factors in some way... can I get some hints?

Note: $$\sin({(m+1)x})\cdot \sin{((m-1)x)}=\\ \frac 12[\cos((m+1)x-(m-1)x)-\cos((m+1)x+(m-1)x)]=\\ \frac 12[\cos(2x)-\cos (2mx)].\\$$ Hence: $$\int_{0}^{\frac \pi2} \sin((m-1)x)\sin(mx)\sin((m+1)x)dx=\\ \int_{0}^{\frac \pi2}\frac 12[\cos(2x)-\cos (2mx)]\sin(mx)dx=\\ \color{red}{\frac12\int_{0}^{\frac \pi2}\sin(mx)\cos(2x)dx}-\color{blue}{\frac 12\int_{0}^{\frac \pi2}\sin(mx)\cos (2mx)dx}=\\ \color{red}{\frac12 \int_{0}^{\frac \pi2}\frac12[\sin(mx+2x)-\sin(mx-2x)]dx}-\\ \color{blue}{\frac12\int_{0}^{\frac \pi2}\frac12[\sin(mx+2mx)-\sin(mx-2mx)]dx}=\color{red}{\frac14\left[-\frac{1}{m+2}\cos{(mx+2x)+\frac{1}{m-2}\cos(mx-2x)}\right]\bigg{|}_0^{\pi/2}}-\\ \color{blue}{\frac14\left[-\frac{1}{3m}\cos{(3mx)-\frac{1}{m}\cos(mx)}\right]\bigg{|}_0^{\pi/2}}=\\ \cdots$$ Can you finish?
HINT: Use formulas for $\sin(\alpha+\beta)$ and $\cos(\alpha+\beta)$ to obtain first what is $\sin x\sin y$.
$$\sin(nx) = \frac{ e^{inx} - e^{-inx}}{2i}$$