0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

HINT: Use formulas for $\sin(\alpha+\beta)$ and $\cos(\alpha+\beta)$ to obtain first what is $\sin x\sin y$.

$\endgroup$
1
$\begingroup$

Hint:

$$\sin(nx) = \frac{ e^{inx} - e^{-inx}}{2i}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.