Integrate $I=\int_0^{\pi} \left(\cos(\theta)\right)^n \cos(p\theta) d\theta$

Question

Is there any standard solution or way to solve the following integration

$$I=\int_0^{\pi} \left(\cos(\theta)\right)^n \cos(p\theta) d\theta$$

where, $n=0, 1, 2,\dots$ and $p=0, 1, 2,\dots$ and $p> n$

Answer 1

Hint:

Using the complex representation with $z=e^{i\theta}$,

$$I=-i\int_1^{-1}\left(\frac{z+z^{-1}}2\right)^n\frac{z^p+z^{-p}}2\frac{dz}z\ =-\frac i{2^{n+1}}\int_1^{-1}\sum_{k=0}^n\binom nk\left(z^{2k-n+p-1}+z^{2k-n-p-1}\right)dz.$$

Then for $m\ne-1$,

$$\int_0^\pi z^mdz=\left.\frac{z^{m+1}}{m+1}\right|_1^{-1}=-\frac2{m+1}$$ for even $m$ and $0$ otherwise, and

$$\int_0^\pi z^{-1}dz=\left.\log z\right|_1^{-1}=i\pi.$$

Logically, all imaginary terms should vanish, leaving either $0$ or $$\binom n{(n+p)/2}\dfrac\pi{2^{n+1}}.$$

Answer 2

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\begin{align} I_{np} &\equiv \int_{0}^{\pi}\cos^{n}\pars{\theta}\cos\pars{p\theta}\,\dd\theta = \half\int_{-\pi}^{\pi}\cos^{n}\pars{\theta}\cos\pars{p\theta}\,\dd\theta:\ ?. \quad \left\lbrace\begin{array}{l} \ds{n \in \braces{0,1,2\ldots}} \\ \ds{p \in \mathbb{Z}} \end{array}\right. \end{align}

\begin{align} \color{#f00}{I_{np}} & = \half\,\Re\int_{-\pi}^{\pi}\cos^{n}\pars{\theta}\expo{\ic p\theta}\,\dd\theta = \half\,\Re\oint_{\verts{z}\ =\ 1}\,\,\pars{z^{2} + 1 \over 2z}^{n}z^{p} \,{\dd z \over \ic z} \\[4mm] & = {1 \over 2^{n + 1}}\,\Im\oint_{\verts{z}\ =\ 1}\, {\pars{1 + z^{2}}^{n} \over z^{n - p + 1}}\,\dd z = {1 \over 2^{n + 1}}\,\Im\sum_{k = 0}^{n}{n \choose k} \oint_{\verts{z}\ =\ 1}\,\,\,{\dd z \over z^{n - p - 2k + 1}} \\[4mm] & = {1 \over 2^{n + 1}}\,\Im\sum_{k = 0}^{n}{n \choose k} 2\pi\ic\bracks{\vphantom{\large A} n -p - 2k + 1 = 1}\quad \mbox{where}\ \bracks{\cdots}\ \mbox{is an}\ \ul{Iverson\ Bracket}\,. \end{align}

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\begin{align} \mbox{Then,}\quad\color{#f00}{I_{np}} & = {\pi \over 2^{n}}{n \choose \pars{n - p}/2}\sum_{k = 0}^{n} \bracks{{n - p \over 2} = k} \\[4mm] & = \left\lbrace\begin{array}{ccl} \ds{{\pi \over 2^{n}}{n \choose \pars{n - p}/2}} & \mbox{if} & \ds{0 \leq {n - p \over 2} \leq n\quad\mbox{and}\quad{n - p \over 2}\ \in\ \mathbb{N}_{\geq 0}} \\[2mm] \ds{0}&&\mbox{otherwise} \end{array}\right. \end{align}

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With $\color{#f00}{\ds{n\ \in\ \braces{0,1,2,\ldots}}}$ and $\color{#f00}{\ds{p \in \mathbb{Z}}}$: $$ \color{#f00}{I_{np}} \equiv \int_{0}^{\pi}\cos^{n}\pars{\theta}\cos\pars{p\theta}\,\dd\theta = \color{#f00}{\left\lbrace\begin{array}{ccl} \ds{{\pi \over 2^{n}}{n \choose \pars{n - p}/2}} & \mbox{if} & \ds{n\ \geq\ \verts{p}\quad\mbox{and}\quad {n - p \over 2}\ \in\ \mathbb{N}_{\geq 0}} \\[2mm] \ds{0}&&\mbox{otherwise} \end{array}\right.} $$