Integrate $\int \cos^n x dx$

Question

I know how to solve it with reduction formula, but is there concrete answer for this integral (without other integrals like in reduction formula)? WolframAlpha gives me expression with $F_1(\frac{1}{2}, \frac{n+1}{2};\frac{n+3}{2}; \cos^2(x))$, but I don't know really what it is. I want to know if there is a simpler solution for this without other integrals like in reduction formula and difficult things like $F_1(\frac{1}{2}, \frac{n+1}{2};\frac{n+3}{2}; \cos^2(x))$?

Answer 1

Using $z=e^{ix}$ we have

$$i2^n\int \cos^nx\,dx=\int(z+z^{-1})^n\frac{dz}{z}=\int\sum_{k=0}^n\binom nkz^{2k-n-1}dz=\sum_{k=0}^n\binom nk\frac{z^{2k-n}}{2k-n},$$ with a special case when $2k=n$ (giving a term in $\log z$). We can regroup the terms in pairs of opposite exponents.

For odd $n$, this is

$$2i\sum_{k=(n+1)/2}^n\binom nk\frac{\sin(2k-n)x}{2k-n},$$

and for even $n$,

$$2i\sum_{k=n/2+1}^n\binom nk\frac{\sin(2k-n)x}{2k-n}+\binom n{n/2}ix.$$

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For concrete examples,

• $n=3$,

$$\int\frac{(z+z^{-1})^3}{2^3}\frac{dz}{iz}=\int\frac{z^2+3z^0+3z^{-2}+z^{-4}}{8i}dz=\frac1{8i}\left(\frac{z^3-z^{-3}}3+3(z-z^{-1})+6\log z\right)\\ =\frac{\sin3x}{12}+\frac{3\sin x}{4}.$$

• $n=4$,

$$\int\frac{(z+z^{-1})^4}{2^4}\frac{dz}{iz}=\int\frac{z^3+4z+6z^{-1}+4z^{-3}+z^{-5}}{16i}dz=\frac1{16i}\left(\frac{z^4-z^{-4}}4+2(z^2-z^{-2})+6\log z\right)\\ =\frac{\sin4x}{32}+\frac{\sin2x}{4}+\frac38x.$$

Answer 2

if n is odd

$\int \cos^{2k+1} x dx\\ \int \cos x (\cos^2 x)^k dx\\ \int \cos x (1-\sin^2 x)^k dx\\ u = \sin x,\ du = \cos x\ dx\\ \int (1-u^2)^k du\\ \int \sum_\limits{i=0}^k {k\choose i} (-1)^i u^{2i} du\\ \sum_\limits{i=0}^k {k\choose i} (-1)^i \frac {\sin^{2i+1} x}{2i+1}$

if n is even...$n = (2^p)k$ where k is odd.

$\int \cos^{2^pk} x dx\\ \int \frac 12 (1+\cos^{2^{p-1}k} 2x)\ dx\\ \int \frac 12 (1+\frac 12(1+\cos^{2^{p-2}k} 2x)\ dx\\ \int \frac 34 + \frac 14 \cos^{2^{p-2}k} 4x\ dx\\ \int 1-\frac 1{2^p} + \frac 1{2^p}\cos^k (2^p x)\ dx\\ (1-\frac 1{2^p})x + \frac 1{2^{2p}}\sum_\limits{i=0}^k {k\choose i} (-1)^i \frac {\sin^{2i+1} 2^px}{2i+1}\\ $

Answer 3

hint

if $I_n$ is your integral,

use by parts integration to find a recursive formula between

$I_{n+2}$ and $I_n$.

Answer 4

Utilize the expansions

\begin{align} &\cos^{2m}x= \frac1{2^{2m}}\binom {2m}{m}+\frac1{2^{2m-1}}\sum_{k=1}^{m} \binom {2m}{m-k} \cos2kx\\ &\cos^{2m+1}x= \frac1{2^{2m}}\sum_{k=0}^{m} \binom {2m+1}{m-k} \cos(2k+1)x\\ \end{align} and then integrate piecewise.