# integral

• Feb 11th 2011, 10:10 PM
dapore
integral
• Feb 11th 2011, 10:13 PM
Chris11
What have you tried so far, have you tried any identities?
• Feb 11th 2011, 10:52 PM
Prove It
Hint:

$\displaystyle \cos{(n\theta)} \equiv \sum_{k=0}^{n}{n\choose{k}}\cos^k{\theta}\sin^{n-k}{\theta}\cos{\left[\frac{1}{2}(n-k)\pi\right]}$
• Feb 11th 2011, 11:13 PM
dapore
• Feb 11th 2011, 11:59 PM
TheCoffeeMachine
Either using $\int_{a}^{b}f(x)\;{dx} = \int_{a}^{b}f(a+b-x)\;{dx}$ or putting $x \mapsto x-\pi$ gives:

\displaystyle \begin{aligned} I & = \int_{0}^{\pi} \cos{x}\cos{3x}\cos{5x} \\& = \int_{0}^{\pi} \cos(\pi-x)\cos(3\pi-3x)\cos(5\pi-5x) \\& = -\int_{0}^{\pi} \cos{x}\cos{3x}\cos{5x}. \end{aligned}

Hence $2I = 0$, thus $I = 0$.

More generally, if $n$ is even, then:

\displaystyle \begin{aligned} & \int_{0}^{\pi}\prod_{0\le k \le n}\cos\left[(2k+1)x\right]\;{dx} \\& = \int_{0}^{\pi}\prod_{0\le k \le n}\cos\left[(2k+1)\pi-(2k+1)x\right]\;{dx} \\& = -\int_{0}^{\pi}\prod_{0\le k \le n}\cos\left[(2k+1)x\right]\;{dx}. \end{aligned}

Thus $\displaystyle \int_{0}^{\pi}\prod_{0\le k \le n}\cos[(2k+1)x]\;{dx} = 0.$
• Feb 12th 2011, 04:56 AM
Krizalid