<s>[INST] Integrate $\int \cos^n x dx$ 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 $F_1(\frac{1}{2}, \frac{n+1}{2};\frac{n+3}{2}; cos^2(x))$?
 [/INST] 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}\\
$
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