<s>[INST] Interesting way to evaluate $ \int \cos^3 x\ dx$ 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?
 [/INST] Use that $\cos 3x = 4\cos^3 x - 3\cos x$ to get:
$$\int \cos^3 x \,dx = \frac{1}{4}\int\left(\cos 3x +3\cos x\right)\,dx$$
and the right side is easy to compute as $\frac{1}{12}\sin 3x +\frac{3}{4}\sin x$.

My original answer included a second approach, but I had the wrong formula for $dx$ in the substitution.
Use the tangent-half-angle substitution, $t=\tan(x/2)$ so $dx = \frac{2}{1+t^2}\,dt$ (corrected) and $\cos(x)=\frac{1-t^2}{1+t^2}$.
This reduces to:
$$\int \frac{2(1-t^2)^3}{(1+t^2)^4}\,dt$$
But this is a bit harder to integrate than the nice formula I had before. You might still solve this if you rewrite $1-t^2=2-(1+t^2)$ and expand, and then compute:
$$\int \frac{dt}{\left(1+t^2\right)^k}$$
for $k=1,2,3,4.$
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