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: $$f'= 6\sin(x) -9\int f(x)dx$$ Therefore,
$$\int f(x)dx=\frac 23 \sin(x) - \frac {f'} 9$$
Do you know any other way to calculate easily this integral?
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