TITLE: Interesting way to evaluate $ \int \cos^3 x\ dx$
QUESTION [11 upvotes]: I have read these days a nice way for integrating $\cos^3x$:
First differentiate:
$$f=\cos^3(x)$$$$f'=-3\cos^2(x)\sin(x)$$$$f''=6\cos(x)\sin^2(x)-3\cos^3(x)$$$$f''=6\cos(x)(1-\cos^2(x))-3f$$$$f''=6\cos(x)-6\cos^3(x)-3f$$$$f''=6\cos(x)-9f$$
Then integrate:
$$f'= 6\sin(x) -9\int f$$
Therefore,
$$\int f=\frac 23 \sin(x) - \frac {f'} 9$$
Do you know any other way to calculate easily this integral?

REPLY [3 votes]: Simpson approach...
$$ g =\int \cos^3(x)dx = \int \cos(x) \cos^2(x)dx$$
$$ g = \frac 12\int \cos(x)  (\cos(2x)+1)dx$$
$$ 2 g = \int \cos(x)  \cos(2x)dx + \int \cos(x) dx$$
$$ 2 g = \int \frac 12 (\cos(x)  + \cos(3x))dx + \sin (x) $$
$$ g = \frac 1 {12} \sin(3x) + \frac 34 \sin (x) + k$$