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?
 A: Euler's approach:
\begin{align}\int\cos^3(x)~dx&=\int\left[\frac{e^{ix}+e^{-ix}}2\right]^3~dx\\&=\int\frac{e^{3ix}+3e^{ix}+3e^{-ix}+e^{-3ix}}8~dx\end{align}
At this point, either integrate directly, and then notice the resulting sine functions, or notice that
$$\frac{e^{3ix}+e^{-3ix}}8=\frac14\cos(3x)\\\frac{3e^{ix}+3e^{-ix}}8=\frac34\cos(x)$$
And thus the rest is simple.
A: 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 $k=1,2,3,4.$
A: \begin{eqnarray*}
\int \cos^3(x) dx &=& \int (1-\sin^2(x)) \cos(x) dx = \int \cos(x) dx - \int \sin^2(x) \cos(x) dx \\
\int \cos(x) dx &=& \sin(x) + C_1
\end{eqnarray*}
Now, we need to integrate $\sin^2(x) \cos(x)$. We do this with the substitution $u = \sin(x)$.
\begin{eqnarray*}
\int \sin^2(x) \cos(x) dx &=& \int u^2 du = \frac{u^3}{3} + C_1 \\
\int \sin^2(x) \cos(x) dx &=& \frac{\sin^3(x)}{3} + C_1 \\
\int \cos^3(x) dx &=& \sin(x) - \frac{\sin^3(x)}{3} + C \\
\end{eqnarray*}
A: A quite different approach would be to use integration by parts. Rewriting $\cos^3x$ as $(\cos^2x)(\cos x)$ and choose $f=\cos^2x$ and $g'=\cos x$, then integration by parts gives $\int{\cos^3}dx=\sin x\cos^2x+2\int{\cos x\sin^2x}dx$. When you replace $\sin^2x=1-\cos^2x$ you get another $\cos^3x$ term behind an integral sign. This way you can solve an equation in terms of $\int{\cos^3x}dx$.
A: Simpson approach...
$$ g =\int \cos^3(x)dx = \int \cos(x) \cos^2(x)dx$$
$$ g = \frac 12\int \cos(x)  (\cos(2x)+1)dx$$
$$ 2 g = \int \cos(x)  \cos(2x)dx + \int \cos(x) dx$$
$$ 2 g = \int \frac 12 (\cos(x)  + \cos(3x))dx + \sin (x) $$
$$ g = \frac 1 {12} \sin(3x) + \frac 34 \sin (x) + k$$
A: Maybe you'll like this method:
\begin{eqnarray}
\int\cos^3xdx&=&\int\cos^2xd\sin x\\
&=&\int(1-\sin^2x)d\sin x\\
&=&\sin x-\frac13\sin^3x+C.
\end{eqnarray}
