Question

integral sen(x+pi/2) sen(2x+pi/2)sen(3x+pi/2)dx

Answer 1

É dada a integral:

$\displaystyle I=\int_{\pi}^{-\pi} \sin\left(x+\dfrac{\pi}{2}\right)\sin\left(2x+\dfrac{\pi}{2}\right)\sin\left(3x+\dfrac{\pi}{2}\right)\, dx$

Usando que $\sin\left(\theta+\dfrac{\pi}{2}\right)=\cos(\theta)$, obtemos:

$\displaystyle I=\int_{\pi}^{-\pi} \cos(x)\cos(2x)\cos(3x)\, dx$

Agora, vamos desenvolver o integrando, usando que $cos(p+q)+\cos(p-q)=2\cos(p)\cos(q)\\\Longrightarrow \cos(p)\cos(q)=\dfrac{1}{2}[\cos(p+q)-\cos(p-q)]$:

$\underbrace{\cos(x)\cos(2x)}\cos(3x)=\dfrac{1}{2}[\cos(x+2x)+\cos(x-2x)]\cos(3x)\\\\ \cos(x)\cos(2x)\cos(3x)=\dfrac{1}{2}[\cos(3x)+\cos(-x)]\cos(3x)\\\\ \cos(x)\cos(2x)\cos(3x)=\dfrac{1}{2}[\cos(3x)+\cos(x)]\cos(3x)\\\\ \cos(x)\cos(2x)\cos(3x)=\dfrac{1}{2}\cos(3x)\cos(3x)+\dfrac{1}{2}\cos(x)\cos(3x)\\\\ \cos(x)\cos(2x)\cos(3x)=\\\dfrac{1}{2}\cdot\dfrac{1}{2}[\cos(3x+3x)+\cos(3x-3x)]+\dfrac{1}{2}\cdot\dfrac{1}{2}[\cos(x+3x)+\cos(x-3x)]$
$\cos(x)\cos(2x)\cos(3x)=\dfrac{1}{4}[\cos(6x)+\underbrace{\cos(0)}_{1}]+\dfrac{1}{4}[\cos(4x)+\underbrace{\cos(-2x)}_{\cos(2x)}]\\\\ \cos(x)\cos(2x)\cos(3x)=\dfrac{1}{4}[\cos(6x)+\cos(4x)+\cos(2x)+1]$

Então, a integral que queremos equivale a:

$\displaystyle I=\int_{\pi}^{-\pi} \sin\left(x+\dfrac{\pi}{2}\right)\sin\left(2x+\dfrac{\pi}{2}\right)\sin\left(3x+\dfrac{\pi}{2}\right)\, dx\\\\ I=\int_{\pi}^{-\pi}\dfrac{1}{4}[\cos(6x)+\cos(4x)+\cos(2x)+1]\,dx\\\\ I=\dfrac{1}{4}\int_{\pi}^{-\pi}[\cos(6x)+\cos(4x)+\cos(2x)+1]\,dx\\\\ I=\dfrac{1}{4}\left[\int_{\pi}^{-\pi}\!\!\!\!\!\cos(6x)\,dx+\int_{\pi}^{-\pi}\!\!\!\!\!\cos(4x)\,dx+\int_{\pi}^{-\pi}\!\!\!\!\!\cos(2x)\,dx+\int_{\pi}^{-\pi}\!\!\!\!\!dx\right]$
$\displaystyle I=\dfrac{1}{4}\left[\left(\dfrac{1}{6}\sin(6x)\right)_{\pi}^{-\pi}+\left(\dfrac{1}{4}\sin(4x)\right)_{\pi}^{-\pi}+\left(\dfrac{1}{2}\sin(2x)\right)_{\pi}^{-\pi}+\left(x\right)_{\pi}^{-\pi}\right]\\\\ I=\dfrac{1}{24}(\sin(6x))_{\pi}^{-\pi}+\dfrac{1}{16}(\sin(4x))_{\pi}^{-\pi}+\dfrac{1}{8}(\sin(2x))_{\pi}^{-\pi}+\dfrac{1}{4}(x)_{\pi}^{-\pi}$
$I=\dfrac{1}{24}(\underbrace{\sin(6(-\pi))}_{0}-\underbrace{\sin(6\pi)}_{0})+\dfrac{1}{16}(\underbrace{\sin(4(-\pi))}_{0}-\underbrace{\sin(4\pi)}_{0})+\\+\dfrac{1}{8}(\underbrace{\sin(2(-\pi))}_{0}-\underbrace{\sin(2\pi)}_{0})+\dfrac{1}{4}\underbrace{((-\pi)-\pi)}_{-2\pi}\\\\ I=\dfrac{1}{4}(-2\pi)=-\dfrac{\pi}{2}\\\\ \boxed{\displaystyle \int_{\pi}^{-\pi} \sin\left(x+\dfrac{\pi}{2}\right)\sin\left(2x+\dfrac{\pi}{2}\right)\sin\left(3x+\dfrac{\pi}{2}\right)\, dx=-\dfrac{\pi}{2}}$